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There come times, in the course of human events, whereby an official-unofficial OTI Staff Scientist such as myself feels compelled to write about pop-cultural issues which he or she (*statistically: he*) finds irksome on a professional level. *Back to The Future* is a nearly endless font of these issues for me—from the obligatory complications surrounding causality in time-travel paradoxes to hoverboards which don’t work above water to the continued reminder of the sheer mathematical improbability of Huey Lewis and the News. But as my esteemed colleagues have already touched on these with great aplomb, I’ve got some another problem to Overthink™ today.

And oh yes, there will be math involved.

Pop culture has envisioned a multitude of time-travel devices, and these give rise to the usual scientific musings (physical and metaphysical alike): questions of causality, internal paradoxes, and the “inconveniences” associated with the creation a billion alternate universes or self-referential circular time lines. Also, there’s usually lightning and dramatic music.

However, there’s a much smaller–though fundamentally difficult–caveat to time travel that’s always kinda’ bothered me. With the notable exceptions of Doctor Who and *Bill and Ted’s Excellent Adventure*, the spatial aspects of time travel are almost universally overlooked when characters decide to go gallivanting through temporal mayhem. I was reminded when questioned by a fellow OTI™ writer in a recent email; so as to protect his/her (*statistically: his*) anonymity, I’ll keep the author’s identity a secret :

**Q.**

*“Watthew Mrather” writes: The DeLorean can’t go anywhere in space, except in the conventional driving/flying sense. Your *[sic]* always in the same place you left, but in a different time.*

**A.**

Well,* *for starters, that’s not a question, “Mr. Mrather,” but I’ll go along with it anyway. Remember, though: without proper punctuation and grammar, we are no better than the animals.

In fact, in order to pull off the kind of time travel we see in the *Back To The Future* trilogy–the kind where the traveler is transposed in time, but remains stationary in the same *relative *position to where he/she left–the DeLorean would have to be an **outstanding** space ship, in addition to its already laudable work as a time-ship. A major issue of freely traveling within time while limiting one’s self to a local reference frame–say, a California mall parking lot–is that the reference frame itself isn’t stationary. As an illustration, let’s figure out how far the DeLorean would have to travel in order to stay in sync with the Earth over a relatively small time-jump. We’ll look at the simplest example (and the first one, diagetically speaking) of the whole *BTTF* trilogy. You all remember the scene, right? (Spoiler alert: Professor Plum and Alex P. Keaton send a dog one minute into the future.)

http://www.youtube.com/watch?v=BytKSy8M4bk

*(Why no one in sci-fi movies ever names their pet Smoluchowski, I’ll never know.) *

As most of us know, the Earth both rotates about its axis (accounting for the regular cycle of day and night, and hence a good bit of poetry) and revolves around the sun (accounting for the seasons, and hence the rest of poetry). However, the litany of additional factors affecting Earth’s absolute position in space makes it fantastically difficult to calculate where a given point on its surface would be after some interval of time has passed. A little professional lingo, here: we in the science biz refer to this level of complexity as a **“colossal mind-fuck.”** So, to make my life easier, I’ll first lay down the standard boilerplate warning we employ for all Overthought(™-pending) mathematical meta-transmutations:

**For all calculations pertaining to a period of time less than one hour, the Overthinking It™ writers reserve the right to disregard effects due to the nonperpendicular axial-tilt of the Earth relative to its plane of revolution, ****the precession of the Earth about its axis,** **the gradual decay of Earth’s orbit into the sun, the** **rotation of the Milky Way,** **analogous rotational and linear velocities within the Virgo Supercluster, and the general expansion of the Univese.**

**Also, we won’t factor in Daylight Savings Time.**

Got it? Good. Let’s get this party started.

SO, according to Doc Brown’s stopwatch, Einstein (the dog, not the Princetonian–or the grossly mispronounced Russian film director) travels precisely one minute into the future on this first jump, arriving, relative to their frame of reference, at the same location he left. But how far has this reference frame *itself* traveled during that one minute? Let’s calculate and see.

**PART THE FIRST. In Which the System Is Set Up:**

We know that *BTTF* takes place somewhere in California–and for the sake of argument, we’ll say that Hill Valley is near Los Angeles, hence at ~35^{o} N latitude. It’s October 26, meaning that it’s autumn in the northern hemisphere, and so for our calculation we’ll take the Earth’s distance from the sun to be roughly perihelion, and unchanged during the minute time jump.

According to Doc’s watch, the time-jump starts at 1:20A.M. Now, the Earth rotates 360^{o} every 24 hours, a rate of 15^{o} per hour. If we assume that Hill Valley was furthest away from the sun at midnight local time, it’s therefore rotated (counter-clockwise, as viewed peering down from the North Pole) exactly 20^{o} since midnight. Here’s a summary of what’s going on at the moment Einstein makes his temporal leap:

**PART THE SECOND. In Which Rotation is Considered:**

As stated above, the earth makes a full rotation (360^{o}) every 24 hours. This is (24hr)*(60 min/hr) = 1440 minutes in a day. Hence, over the course of a single minute, the earth will rotate (360^{o})/1440 = 0.25^{o}. The Earth’s radius at the equator is 6378.1km, but with increasing latitude the distance between a surface observer and the axis of rotation decreases (consider the case of a person standing on the north pole–he/she *(statistically: he)* wouldn’t move at all due to rotation. Note also: for subterranean Mole-Men, the calculation’s even harder.) But a little trigonometry tells us that the distance between a surface observer this axis is (6378.1) * cos(latitude), or (6378.1)*cos(35) = 5287.1km.

So the point at which Einstein takes off/lands is at a distance of ~5300km from the Earth’s rotation axis, and rotates through a 0.25^{o} angle during the minute. Checkout figure two, which summarizes the effects due **only to rotation**:

Now, in order to solve for D_{1}, we’ll take the standard OTI approach and–just as Mr. Belinkie did for the *Dark Bailout* movie–employ the law of cosines. To jog your memories:

Plugging in a=b=5287.1km, and γ =0.25^{o}, we arrive at…

D_{1} = 23.07km

Also, (though this isn’t labeled in figure 2), since the triangle in Figure 2 is isosceles, each of the non-0.25^{o} angles are equal. And since the sum of the angles of a triangle is always 180^{o},these therefore have a value of 89.875^{o}.

**PART THE THIRD. In Which Revolution is Considered:**

What about the Earth’s revolution around the sun? Here’s what we know. Since the Earth is ~perihelion, we’ll call the distance between its center of mass (the Eternal Winter Palace of the High Czar of the Mole-Men) and the sun to be 149.5 million kilometers. In truth, our orbit is an ellipse, but it’s not *that* elliptical *(I mean, I’ve certainly seen more elliptical ellipses–like this: “…”)* so we’ll estimate it as a circle. To fully traverse this circle the Earth takes a full year, which is 365.25 days. Hence, despite anything certain overtly melodramatic people might tell you, a year consists of:

(365.25 days/year) * (24 hours/day) * (60 minutes/hour) = **525960 minutes.**

(How do I measure, measure a year? *With some fucking precision.*)

So as before, we calculate that in a minute’s time, the Earth revolves 360^{o}/525960 = 0.0006844626^{o}. Check out figure three which takes into account the effects due **only to revolution**:

Again, we’ve got an isosceles triangle to which we can apply the law of cosines to yield:

D_{2} = 1785.88km

And as before, the two larger angles within the triangle are calculated to be 89.99966^{o}.

**PART THE FOURTH. In Which We Finally Get to Go Home:**

Let’s look at the overall effect *en toto*. Though revolution and rotation are of course simultaneous, we can parse the mathematical effects of both out into separate steps. Hence each displacement becomes the leg of a triangle–one for rotation, one for revolution. Check out figure four–which **puts it all together:**

We already have D_{1} and D_{2}. From the calculations in figure 3, we compute the smaller of the two interior angles of the central parallelogram to be 180^{o} – (89.99966^{o} + 20^{o}) = 70.00034^{o}, as is shown in Fig. 4A. From Figure 2, we know that the small isosceles triangle on the right of 4A has angles of 89.875^{o}. As in figure 4B, adding these two gives a total angle of 159.87534^{o}. Applying the law of cosines one last time yields:

**D _{3}=**

**The total distance Einstein would have to travel in order to arrive back in the Twin Pines Mall parking lot one minute into the future…**

**=1807.56km**

**=1123.17miles**

**(!!!)**

That distance is ~4/5 the length of the eastern sea board of the United States! Or–to put it more juvenile terms–it’s approximately the distance from Bangor, Maine to Morehead, Kentucky. But less sexy. And that’s how far the Earth would travel in just one minute! I don’t even want to go near the calculation for the 1985-1955 jump…

Now, the machine *is* traveling at 88mph. While that’s hardly fast enough to cover this distance, how fast would it have to go to make the jump in space as well as time?

Speed is just the distance traveled over an interval of time. Doc’s stopwatches would have you believe that Einstein’s trip took *no* time at all, at least as far as the dog is concerned:

But then, his watches aren’t *perfect*, and this allows us to try and estimate something we’re unable to measure. Let’s be generous and assume that the stopwatches have an error of ~1 millisecond. That is, Einstein’s watch could be .001s slower than we think it is, without a perceptible effect: we’d still see the two watches changing time in pretty much perfect sync with one another. Hence an upper-limit to the time it took Einstein to jump 1807.56km is 0.001s. This yields an average speed of:

(1807560m)/(0.001s) = 1.808 x 10^{9} m/s

…roughly **six times the speed of light in a vacuum**. Of course, an object moving faster than the speed of light would theoretically travel ** backward **in time, so that kinda’ bones us right there.

* * *

So what can we take from this? Well first off, were the DeLorean to really just pop-out and pop-in to time in the same absolute place, then even this first tiny future-leap would drop poor Einstein nearly two-thousand kilometers into the vacuum of space. Longer leaps of time would be even more potentially disastrous: you might reappear inside solid rock, or within the fusion furnace of a sun, or near Steven Segal. One shivers at the thought.

But that kind of speculation is moot. We know that the DeLorean *does* pop back in pretty much the same relative place it popped out. Maybe it’s a conscious choice–maybe it’s a nimble space ship that Doc could have engineered it to go anywhere, but kept it within frame for safety’s sake. After-all, while I’m unable (and certainly unwilling) to do more elaborate calculations, that’s not to say that –given more sophisticated calculation equipment–a computer on board the DeLorean wouldn’t be able to handle them with aplomb. I mean, check out all the buttons and toggle-switches inside that thing–maybe some of them control the distance-computation circuits, in addition to the flux capacitor *et. al.* This gets less and less plausible when one imagines how Doc Brown could have constructed a second time-ship out of a 19th century train–as is revealed in the closing scene of *BTTF 3*. After-all, the retroengineering required in fashioning a flux capacitor from pre-period goods would be hard enough without having to fabricate silicon microprocessors as well.

No, my guess is that we’re supposed to assume the DeLorean is somehow coupled to the local reference frame as it travels freely through time–much in the same way that you and I are coupled to it as we travel through time in a constrained fashion. While the easiest way to achieve this coupling is through gravity, the big “G” tends to muck up time in whatever way it sees fit, regardless of your fancy-shmancy time travel plans.

Now that I think of it, one could always synthesize a worm-hole that encloses a timeless space, with its exterior coupled to the Earth’s timeline. This is of course absurd: any object passing through it would be exposed temperatures approaching absolute zero and would hence instantly become *incomparably cold*.

Hrm.

mlawski#Man, I sure hope the DeLorean time machine also takes into account leap seconds. Otherwise, they are SCREWED.

Matthew Belinkie#@Dave – Well done! I know you said in your disclaimer you weren’t factoring in any movement beyond the solar system level. But I just can’t help myself – can you give me a rough estimate about how fast the universe is expanding and the Milky Way is rotating? It’s got to be a tremendous speed, right?

Matthew Wrather#1) Mea culpa. I made a typo.

2) I believe a year is 365.24 days, not 365.25 (which is why the last year of every century is not a leap year, despite being divisible by 4).

I point this out because without precision and exactitude in our measurements, we are no better than the animals.

Lynsey#I just assumed anything in earths gravitational field at point a and point b would be static regardless of the axis spinning. I’m too simple!

Nathan Miller#It seems to me that there’s relativity forgotten here. I mean, by this logic, when I jump up, I’d have to jump at nearly the speed of light to land a few seconds later. That’s ludicrous.

Instead, when I’m driving a car, the speed of my car is 50 mph, _relative to the ground_, not relative to the sun. In reality, I’m traveling just as fast as the earth is, plus or minus up to 50 mph (depending on my direction, obviously).

Similarly, the DeLorean, when it hits its speed of 88, is already accelerating (because of the turn) at the necessary rate to catch up with the Earth, just as my Chevy does when it reaches 50. The result is that the car, when it hits a minute later, is still going the original 88 mph, PLUS the speed/acceleration of the Earth, necessary to catch up with the current location of the Twin Pines Mall (soon to be the Lone Pine Mall, of course.)

More importantly, it’s just a fun movie.

girlbehindtheglasses#Fabulous discussion here. I remember watching all of these movies in a row one night with my friends. I’m a person that tends to over think things myself, but I’d mostly succeeded in holding back. It was the end of the movie, however, with the train coming in to land and I finally announced. “At least the car kind of made sense.” My friends’ were silent for a moment before turning in unison to stare at me as one of them asked, “That’s your problem with this?” “It’s a train! How the hell do you plan to park a train in the future?” They proceeded to break down into giggles and refused to discuss it, thinking me a silly person for point it out. Now I’m sending them this.

Gab#girlbehindtheglasses: They probably thought you were channeling Captain Obvious.

Barring the temperature thing, could you give a better description of how the wormhole would work? What do you mean by “coupled”? Do the two ends attach themselves to things instead of just hanging there in open space?

fenzel#@Nathan Miller

I’m no scientist, but I think you’re failing to take into account two things

1. Angular momentum – the DeLorean isn’t traveling in the same arc that the earth is – instead it might shoot off in a tangential line to the orbit of the earth, modified by its rotation

2. The DeLorean skips the intervening seconds, which screws up its position relative to the Earth.

Even if the DeLorean is traveling in the same direction as the Earth (which it isn’t) and the same velocity as the earth (which it isn’t) when it time travels, and even if it’s trying to get to the same place as the Earth, it has a lot less time to get there, which means it would have to go a lot faster.

It’s the difference between braking your car to a stop or slamming into a tree – the difference in time frame of the stop may seem somewhat trivia to some human observers (half a second vs. much less, but still within half a second), but the energy is going to be transfered a lot faster to the driver.

Reverse that – try to get from 0 to 60 in a millisecond rather than a few seconds – and you need to generate an immense amount of extra power.

hoof#So up until now, we assumed the delorian disappeared from the universe as is, and appeared in the universe in the near future. Cute, but how about this:

As any good internet junkie knows, time is a dimension, in which we are flowing in one direction at a some-what fixed rate. The delorian just hops ahead in this 4th dimension, following the path of the earth through the other 3 dimensions, and taking the “1 second later” exit, to arrive in the future state of the earth. To go into the past, it takes the southbound exit on the 4th dimension highway, and follows the 3 dimensions of the earth until it finds its exit. Sounds crazy, but if you follow the “ant on the table” analogy, it’s at least believable with a little mind stretching.

shechner#Hey, everyone! Thanks for taking an interest in my little uber-gedanken(tm) experiment.

@Nathan Miller – you raise an excellent point, but I think Fenzel’s responses to you are quite good. Let me rephrase what he’s said so as to respect your obeisance to Relativity. I’ll touch on his second point first, and then throw some math behind his first point.

2) What this article effectively calculates is a displacement vector – really, just how far has this point on the surface of the Earth moved, and in what direction relative to its starting point. This, of course, is all relative to the time/reference frame that Doc and Marty experience – the one in which this distance is traversed along that path (in net) and over the course of a minute.

Now if we put ourselves in the dog’s reference frame, we’ve (again, in net) traversed this same distance, but it’s taken us ~0.0001 seconds to do it. As far as Einstein’s experienced it, he’s traveled a tremendous distance in an extraordinarily small period of time, hence giving him an apparent velocity (as far as the Newtonians would have it) of ~6x the speed of light.

The point, then, is to say that the Newtonian interpretation could not possibly work for a machine like this, and to illustrate it in relatively (pun intended) simple terms. Einstein is *of course* not popping out- and popping back in to the same reference frame, because the result would (as I said), drop him in the middle of space.

1) Or would it? As you pointed out, at the moment the dog Marty/Doc’s reference frame, he’s not traveling at 88mph, he’s traveling at 88mph relative to the Earth. BUT, as Fenzel pointed out, the Earth isn’t moving in a linear fashion, it’s rotating and revolving. These are examples of circular motion – whereby at any instant in time, the velocity vector a body experiences is perpendicular to the radius between it and the center of rotation. That is, for the earth’s rotation, Einstein’s velocity vector is tangential to the earth’s surface. For revolution, the vector is perpendicular to the “to the sun” arrow in figure 1. At the point Einstein “drops off the grid,” the result on his velocity would be roughly equivalent to what would happen if you swing a rock tied to a string around in a circle (something intro-physics textbooks seem to assume we do all the time, for some reason) and then let it go. It ceases to swing in a circle, and continues along the vector path it had in the moment it was let go.

So, Einstein’s total velocity vector at the time he leaves on his time jump is the vector sum of the two velocities due to rotation and revolution. This is a synthesis pretty much *exactly* like the one I walked through in this post, but since we’re not taking about mean displacement over time, but rather instantaneous velocities, we’d have to do the calculation somewhat differently. Namely:

Rotation: we know that at 35 degrees N, a point on the surface would rotate through a full circumference (2*pi*r) in 24 hours. This gives a velocity of ~1384.16km/hr, or ~57.67km/min, perfectly tangential to the surface.

Revolution, again, the total path length (overlooking ellipticity) is 2*pi*r, but the r now is the distance from the earth to the sun. The same math gives us an instantaneous velocity of ~107159.7km/hr, or ~1786.00km/min. However, a little geometry tells us that – at 1:20am – this vector is angled 20 degrees away from the Earth’s surface, aimed skyward.

The final vector sum gives us an instantaneous velocity of ~1764.33km/min, angled at ~19.7437804 degrees from the Earth’s surface.

Our displacement vector (what we calculated in this article) is ~1807.56km, and angled at ~19.99966 degrees from the Earth’s surface.

NOW, let’s consider two scenarios:

A) Like I said in the article, Einstein pops out of the time frame at 1:20, and pops back in at the same point in space, but one minute later. This STILL puts him into the vacuum of space, ~1807km behind the Earth’s orbit, but we now know that he’s traveling at roughly 1760km/min toward approximately where the earth is going. He’ll never catch up, though, the poor little bastard.

B) Like you intimated – maybe he’s, like, put into “suspended animation” for a minute. So his whole time frame just sits tight for a minute in our time frame, continuing along the vector it established at the moment it popped out. Noting the differences in the velocity and displacement vectors, a little math tells me that Einstein would “pop in” 43.953km off-mark.

And by “off-mark,” I mean he’d be ~27.3 miles under the surface of the Earth’s crust, somewhere off the Pacific coast.

Now, I’m not taking into account the 88mph, because (A) it’s a fairly trivial velocity, given the *literally* astronomic ones imparted by rotation and revolution, (B) we don’t have any information about its direction relative those of rotation and revolution (if he’s heading directly due east, for example, he’s actually making the problem worse by adding to the rotation vector), and (C) even if it *did* make up for the lost 43.953km and 0.2558796 degrees between our velocity and displacement vectors in this one case, we know that 88mph is the magic number the DeLorean needs to make *all* time-jumps of all magnitudes. As the displacement/velocity disparity is going to vary with the time difference, it’d certainly not work in every case.

*deep breath*, okay that’s it for now. I’ll get back to Gab’s, Blink’s, and hoof’s comments in my next one.

Again, thanks for reading!

shechner#@Hoof: Again, an excellent point! Yep, time is a dimension, but that doesn’t give you free reign to disregard the others.

For the non geometrically-savvy, a “dimension” is just another axis you can draw (like on a line-graph), but we put the restraint that it must be perpendicular to any other axis you’ve defined for the system. In our five-dimensional world, time is perpendicular to each of the length, width, height, and umami dimensions.

The problem with time is that – as we perceive it, at least – we do not have free reign to wander through it as we do the others. There are numerous possible explanations why (my favorite involving the second law of thermodynamics), but we’ll leave that discussion there and assume that non-DeLorean-owners must travel along the arrow of time just as it’s been shot.

However, just because Doc and Marty can now move along this new dimension however they see fit doesn’t mean that they can disregard the others. Let’s take a lower order analogy:

Let’s picture a three-dimensional world where everyone’s stuck to time as we are (dimension one), and have free-reign to move throughout the other two dimensions (length and width) as we do. That is, it’s a time-dependent flat universe. Now let’s say a two-dimensional Marty and Doc create a depth-machine that can pop above or below the plane of their existence and go to any depth they see fit.

NOW, let’s say they’re standing at a point in their 2D world (x,y)=(0,0) wanted to travel to an analogous point in their own 2D world (1,1), but raised to some new height (which only they can achieve, given their splendid machine) – that is, (x,y,_z_) = (1,1,1). Just because they can explore this new depth dimension doesn’t mean they don’t *also* have to go along the two traditional dimensions, if they want to get to their proposed destination. That is, just because everyone else can only go to (1,1,0) and they can go to (1,1,1) in their machine doesn’t excuse them from having to cross the distance between (0,0) and (1,1) in the 2D plane.

Now, if we picture this world as being like a flip-book, where every 2D plane captures a moment in time, and “time” is now hashed out in the depth dimension, we end up with the exact same scenario as described above, but Doc/Marty’s “depth-machine” has become a “time machine.” That is, as they travel through time, they must also travel through space in the traditional sense. If their universe rotates/revolves like our world – such that (0,0) moves to (1,1) over the time period they want to jump – their machine would have to traverse all three dimensions, not just time, in order to make it back intact.

The ants-on-a-table analogy is great (especially for Star-Trek fans), but it assumes that the machine generating it can warp both time_and_space. Again, this is pretty much the point of my article – without some way of overcoming the spatial problems, having a time machine is pretty much useless.

-Sheq

Jim Hartley#Is it possible that Doc Brown unknowingly invented the Bergenholm and incorporated it into the Flux Capacitor? It is well known that inertialess matter can travel faster than light, limited only by the resistance of the surrounding medium. Therefore, while the Flux Capacitor moves the DeLorean in time, the Earth, the air, the parking lot, and such simply drag the car along with them, at any FTL speed needed, and it emerges from the time jump in the same relative position from which it left.

(If you don’t know what a Bergenholm is, SHAME ON YOU!)

Vadim Lebedev#The Flux Capacitor surely have chracteristics of Impobability Drive too. So space jumps poses no problems

JimT#The DeLorean freezes the passengers reference frame while it drives on auto navigation a nice 55mph and 55hph until it gets to its final destination. Upon arrival it unfreezes the passengers and it appears that they arrived instantaneously at the new location when in the DeLorean frame it took a very long time to get there. If it goes slow enough in hph, it doesn’t have to worry about even bothering calculating where the Earth was since it will always be in relative contact.

Jeff#Sorry if anyone has mentioned it already, but let’s not forget that the earth’s rotation is actually slowing down very slowly. May not make a huge difference over say, 100 years, but will be a huge difference in a million. Forget going back to see the dinosaurs.

John#What if the action of traversing the dimension of time from N to N + 1 minute took no time in and of itself? Then the earth (or anything for that matter) wouldn’t have moved at all because some amount of time is necessary for movement. So if Einstein popped instantly from now to now + 1 minute along the dimension of time, then would anything at all have moved?

Movement in the 3rd and 2nd dimensions require a duration, which is apparently the purpose of time (to allow movement). What if ‘something else’ besides time is required to enable travel within the dimension of time?

John#Oh, nevermind, I get it. If I were to travel into the future instantaneously, using no time myself to make the journey so that nothing in the 3rd Dimension changed….then I haven’t traveled into the future, because the future IS the sum of the movement of all matter in the universe over time.

Well, in that case, I’ll just stick with the traditional method of future time travel. The one we call waiting. It’s safer and easier.

Marc#@John,

Thats what I was thinking. Maybe some quantum mechanics and physics play a part here. If all possibilities of a universe exist in any given moment it should be easy to move along the branch of one to another within that moment, hence not needing to factor in the calculation of time.

Mike Johnson#Inertia? You “stay” in one spot due to a multitude of forces – so in one minute, your frame of reference might also move in relation due to the inertial mass of the object in the frame… like riding in a van, and jumping up in the air, and then coming down again – this, I think might explain the one minute jump forward, but definitely not what happens going back 30 years.

Colin#Relativity is being ignored here. Aside from the fact that we’re breaking special relativity by having a time machine in the first place, you can always pick a reference frame where the DeLorean will disappear and reappear in the same place. You imply looking at it from the sun’s reference frame, or maybe Virgo’s reference frame. Rest assured, cosmologists agree that there is no center of the universe, because avery single point in the universe shows cosmological expansion. Given the ability to jump in time from one point in a non-inertial reference frame (the Earth’s) to another, there would always exist one where the space coordinates are the same.

shechner#@Colin – absolutely true, but that reference frame would (A) not necessarily be on Earth, and (B) Would *absolutely* not be stationary with respect to it – since the fun aspects of “location” and “simultaneity” only apply for speeds which begin to approach c. So while a reference frame could be chosen to suit your criteria, it’s hard to see how the machine would make use of it, seeing as it wouldn’t be part of that reference frame.

Still, that’s probably the best argument yet against everything I’ve written in this article. =)

As for the multiple universe/timeline people (including Mr. Mark Lee): I’ve never been comfortable with these hypotheses. As a disclaimer – I’m professionally a biochemist, and so my training relies heavily on thermodynamics. According to the first law, if jumping back/forward in time involved leaping into a completely alternative universe, then either:

1) The first law of thermodynamics breaks down during time travel, as the creation of a new universe would require (at least) as much energy as is found in our current universe, or

2) There *already are* literally an infinite number of parallel universes, in each of which every possible combination of events is simultaneously being played out. Energy exchange between these would be strictly forbidden, of course (or in the bare minimum, none would be allowed into our universe). In such a scenario, time travel would just be a lateral leap to an existing alternate place.

I’m not thrilled with either option, but given that we really would have no data from alternate universes, there’s no evidence one way or another to lend/remove plausibility.

No, for my (nearly nonexistent) money, I think the fact that we don’t see time travelers is either due to the limitations of physics itself, or to the limitations of biology. Perhaps no living matter (or nothing as delicate as Earthly biology) could survive the requisite conditions during a time-jump. Or maybe our perception of time itself is so hard-wired into our biology that we cannot perceive those things around us which travel through time in a different manner. Like colors in the UV range, we’re just not equipped to see them…

Again, you are all tremendous for giving my OverThinking such additional thought!

-Sheq

CaptObviousman#With the notable exceptions of Doctor Who and Bill and Ted’s Excellent Adventure, the spatial aspects of time travel are almost universally overlooked when characters decide to go gallivanting through temporal mayhem.I present to you the movie Primer, which addresses this difficulty in a unique way. I’d love to see you guys do an article on it! It was filmed in my hometown by an electrical engineer, so many of the design scenes ring very true for me.

Correa#“Now that I think of it, one could always synthesize a worm-hole that encloses a timeless space, with its exterior coupled to the Earth’s timeline. This is of course absurd: any object passing through it would be exposed temperatures approaching absolute zero and would hence instantly become incomparably cold.”

Well, that explains why it is so cold on re-entry after that first jump. Amazing how worm-holes were so well understud by 1985 movie makers. :)

DeusExMach#I think we’re neglecting Both the fact that the DMC-12’s stainless steel construction allows for proper flux dispersal (Look Out!) and the effects of kismetic probability.

First, we agree that by sending something into the past, or into the future, or, indeed, back to OR from the future, one must move through both space and time. But the frame of reference that one leaves from would (theoretically) undergo IMMENSE space-time strain. If a paradox develops, it would be focused back through that “tear” in subspace/hyperspace/what-have-you, and the resulting “flux” of energy would obliterate one or both space-time-frames. If either the tear is fragile enough, or the paradox is powerful enough, it could destroy the entirety of both universes. Of course we could get lucky, and the damage would be localized to our own galaxy. This is why the week of November 5th-November 12th, 1955, October 26-27th, 1985, and October 21-22, 2015 are such dangerous times and places to be. In a matter of two days in 1985 there are a total of 5 temporal incursions, and 6 temporal excursions. The space-time fabric of that short a period of time is remarkably thin and vulnerable to paradox. This is why there are so many different versions of October 21st-22nd, 1985 (Marty’s parents are losers, Doc gets shot, Marty’s parents are rich, Doc was wearing a bullet-proof vest, Marty’s Dad’s been martyrd, Marty hits the Rolls-Royce, Marty sticks it in reverse, because Flea’s a dick-head, etc. Like-wise, there are three temporal incursions and three temporal excursions on the same day in 2015, but as no one traveled BACK to 2015 from a future-point, only minor changes take place in the 2015 time-line. Now, November 5th, 1955 is the red-letter date in the history of science when Doc was hanging a clock in his bathroom, slipped, hit his head, and when he came to, had a vision; a revelation of the Flux Capacitor. There is no explanation of “How” it works, or “Why” it works, other than that it DOES work, and that Emmett L. Brown finally invents something that works. Intriguingly enough, in the time-line that Marty creates, the Flux Capacitor itself exists BEFORE the idea for its existence has come into being. There are three temporal incursions, and 4 temporal excursions over a period of eleven days. The fact that Old Biff Chooses that time period to travel to, the famous Hill Valley lightning storm occurs which strikes the clock-tower on November 12th, electrifying the clock, the Doc, AND the Time vehicle simultaneously, shortly after ANOTHER bolt of lightning from the same storm strikes ANOTHER Doc, screws up ANOTHER Clock, and electrifies ANOTHER Time Vehicle, sending it back to 1885 may hold some sort of cosmic significance. Of course, it could all just be one great big coincidence.

Where Kismetic Probability factors in is that the trials and tribulations of one Martin “Alex P” McFly could well have been preordained. He ALWAYS was going to go back to 1985 to alter history, thereby creating several varying future time-lines, and was ALWAYS going to travel into the future in order to, by way of perceived threat of paradox, set the timeline in the state that it achieves at the end of the third film. In which case, Kismet renders the concept of paradox simultaneously moot, and all-important, which is a pretty neat trick when you can pull it off.

Basically, the Time machine is made of stainless steel, and requires 1.21 Jiggawatts (not to be confused with Gigawatts. Gigawatts are measures of power output; Jiggawatts are measures of Electric Temporal Flux Capacity) to power the Flux Capacitor, which turns the DMC-12 into a nuclear and/or fusion-powered Electro-magnet. The addition of 88 mph is what is required to move the Time Vehicle’s magnetic field against that of the planet earth, which is another really big electro-magnet, thereby allowing a “Flux Resonance” if you will (and even if you won’t, I think I’ve moved beyond common sense, here) That imprints itself on a specific point on the Earth’s Surface. That “Quantum timey-wimey… STUFF” Degausses when the Time Vehicle re-enters normal space-time. For this reason, only certain locuses in space-time are accessible via the Time Vehicle. For Time travel to be possible, Someone will have to leave from the general vicinity of where you are at some point along your personal timeline. The reason that Doc Brown made it back to January 1st, 1885, and was able to travel to the future to refit the “Time-Train” with a hover-conversion, travel BACK to 1985 with Clara, Joules, Verne and Einstein is three-fold:

1) When the DMC-12 Time Vehicle left 1885, it reached Flux Capacitance along a railway track, making both a longer geographical area, and a longer temporal period time-vehicle accessible (the track through the eponymous “Hills and Valley” of Hill Valley, and the time period of 1Jan1885 to when-the-hell-ever-Locomotive-ELB-left-1885).

2) Kismetic Probability dictated that it would happen.

3) The Power of Love.

SiliconBear#OK, rather than thinking the Delorean ‘travels’ from point A to point B, maybe what the Flux Capacitor does is generate a wormhole instead. That what that bright flash is which the Delorean seems to drive into. The 88mph is simply how fast it has to travel to safely transition through the entry aperture before it collapses.

If this is the case, then the movie suggest that the entry side at least MUST be tied to some local reference (for example, it seems to be stationary relative to the ground, and not travelling along at the same speed as the vehicle which generated it). Given this, perhaps the exit aperture is tied to that same reference. I have no idea what mechanism could be behind that…. other than of course it being FICTION.

Mobius#In my time travel novel, which also assumes (at the start) a constant frame of reference, I use a system which pushes and PULLS through time. The sphere which is sent back in time (a la “Terminator bubble”) is replaced by the sphere which WAS back in time… This means you can’t appear inside a wall, or a rock, or have what happened to the FLY in the last movie. Only Time Travel numbskulls allow their body to become part of something else when teleporting/time travelling.

Also, the process to 2-stage, you disappear from NOW and unless you click the AOK button on your controller within 2 seconds of materialising, you get yanked back automatically. You can retrun to your original time and place by clicking the button on your return device. It sends a signal through time. Don’t ask how. It’s a plot device, so it doesn’t need too much explanation anyway!

Important to remember that time travel IS teleporting. Just set your arrival time at 1/100th of a second from now, and the destination to be somewhere else – away from your frame of reference.

The computers are smart enough to figure it all out. Thank God (TM).

When combined the time travel device is a teleporter as well as a time machine. In later builds it is also a perfect replication device, with materials required to create copies pulled from time itself! (I don’t explore this in the novel though).

I always liked the “Mass Damper” effect in the Larry Niven novels, with stepping disks required to create or damp kinetic force on arrival. This is important – because travelling in time is no good if you’re doing 1200 miles an hour when you materialise! Watch out bystanders.

I’m pretty confident that a human body travelling at MACH speed numbers also makes a fairly good kinetic weapon! Talk about human Jam! Yummo…

Anyway, the system needs a mass damper, and mine is the same as Niven’s; a giant water reservoir with a moving mass at the center, causing giant waves at the outside. I also generate some power from these waves, thereby damping the damping action. Not too sure if any laws of thermodynamics are broken in the process, and I don’t care. It’s my novel, let me have it. Thanks.

Mostly my novel is humour (in the Stainless Steel Rat vein) and relied heavily on the stupidity of English when trying to talk about things which have yet to happen, which happened to you in the future, or which might not happen provided you do your part…

Most importantly, my novel works on the principle that it is impossible to go back in time to murder your Dad before you were born. Time doesn’t work like that, and if you go back in time, and return, you are really fulfilling your destiny, because your world already featured what you did in the past before you even travelled there!

If you tried to kill your Dad, it doesn’t work, or it turned out not to be your Dad.

This brings about a philosophical question about freedom of will – but of course it’s circular: if you decide NOT to go back in time, that’s OK, because in your time line you DIDN’T go back in time… And if you DO go back, that’s OK too, because your time line featured what you did in the past.

The universe doesn’t break as a result.

My novel is about the emergence of the human race; in particular the emergence of Homo Sapiens over Neandethals. And (spoiler) it turns out modern man fathered modern man. Don;t ask me how that works either. Thanks.

DeusExMach#@Mobius: In my version, the time traveller is actually Merlin, who is actually a physicist named Paul Merriden from the mid-21st century. His experiments discover that your own personal timeline is like a big rubber band. Pull it enough (by creating a paradox), and it snaps, sending you rocketing backward through time, erasing your life-story like a VHS tape as you live it, in reverse, in ever increasing chunks of time, until you reach a point where you were never born, but as you’re dislocated from time, you don’t die, your age keeps getting reset until you reach a long enough time in the past where you can FINALLY die of extreme old age, but not before some curious proto-human sees you leaning on a stick to walk, which, with your final demise, encourages the knuckle-dragger to use the stick to stand upright, and imagine a life of walking on two legs for the first time, while using tools.

…am I about close?

Aaron Davies#an average gregorian year is 365.2425 days long ((365*400)+97)/400. the actual period is estimated to be 365.256366 days, though at that point, tropical vs. sidereal year makes a noticeable difference. (i assume a “day” is exactly 86400 standard metric seconds here.) astronomers do tend to work with julian years of 365.25 days.

Dan#Wonderful article!!!! Thank You!! …another interesting approach from Science Fiction to time/space travel was the older show Seven Days… in which the writers may have correctly answered the space travel issues along with the time; with the joystick to control movement. Just a thought!!

Cledrik Buonrij#Thanks, I always thought that as well (about how time machines in fiction seem to magically bind themselves to the same space.)

Also, thanks for ruining my enjoyment of this film. Now I have this knowledge, the suspension of disbelief is impossible.

Erik#so — is it completely out of the question that “time travel” did not actually occur, so much as wormhole travel to a parallel universe where events are delayed by the displacement of time set inside the DeLorean’s computer systems?

This way, the DeLorean actually passes through a “window” into an exactly time-accurate parallel universe — which would appear as if the car instantly traveled an infinite distance. The flux-capacitor would simply allow for the opening of the window to the specific parallel world, and each parallel universe has some sort of calculable equation that allows for specific times to be hit…

This certainly hurts the “Marty’s future in peril” storyline, but it gets rid of the distracting “time travel” issues and focuses on linear “normal” time across successive parallel universes…

Behold the power of the script… I love it!

AStudent#Sorry if someone has pointed this out already, but these calculations are wrong. It is, at least according to Einstein, not possible to know where the dog will end up. It has been shown that the universe is not static, our position in relation to the stars does not stay the same. What I’m saying is that we don’t currently have any idea what direction we’re actually moving in. As the universe has no absolute reference frame, we can not assume that velocities that Shechner uses in his (her?) calculation are correct. Simply put, how can you prove that you aren’t moving ten thousand miles an hour away from some other reference frame and that the dog won’t end up 167 miles from where you say he’ll be after a minute? To do this calculation correctly you would have to violate relativity by assuming the existence of an absolute reference frame. Also, and as an aside, these calculations don’t answer the problem of inertia. If the dog is moving at such and such a speed when he leaves the first frame of reference shouldn’t he still be moving that same speed when he comes out? Making the assumptions that the author makes would leave the dog and car kind of mashed upon re-entry.

shechner#@AStudent:

First off, I’m a “he,” though you’d question it if you saw my basketball skills. :) The rest of your post is quite accurate, though unnecessarily defeatist:

Indeed, the General Theory shows that we cannot distinguish between any non-accelerating systems, regardless of their velocities. However, implicit in my math is that we’re establishing an arbitrary local reference frame, which is perfectly reasonable for any calculations. Indeed, General Relativity doesn’t state that we cannot know velocities, positions, etc – it merely states that the absolute numbers we derive for these characteristics will be relative to the frame of reference we choose.

So – I’ve done away with Hubble-constant-dependent variables for concision (and calculability), and treated everything as intra-solar system. The math creates an arbitrary zero – located at the point on the Earth’s surface at 35 degrees N that is experiencing midnight when Hill Vally is experiencing 1:20AM. Once the arbitrary zero is established, all of the numbers are internally consistent, which is really all that any number can be, given Relativity.

(After all, to say that any number would be different, given a different reference plane, would also discredit things like traffic tickets, projected Airplane arrival times, and whether or not Heath Ledger is dead. What you’re saying is *technically* correct, but of course math which overlooks it will be internally consistent, given certain criteria.)

Speaking of said criteria, I’ll point out that relativistic phenomena *only have appreciable effects for systems traveling near the speed of light.* Since nothing in my calculations (save for the disproof-by-null-hypothesis where I calculate the DeLorean’s effetive Newtonian “speed”) are travelling anywhere NEAR 0.001c, relativistic effects are moot on them. Indeed, an observer travelling at, like, 0.8c, passing right by the Earth as Einstein made his jump would see things differently, but no one else would.

Oh, and as for the intertial calculations, check out my previous responses above. Ein’d be going at ~1750km/min, at an angle of ~19.75degrees elevated above a tangent line at his point of departure. It’s a good point, though!

Now, back to my job as a biochemist. :)

John V#@fenzel

“Reverse that – try to get from 0 to 60 in a millisecond rather than a few seconds – and you need to generate an immense amount of extra power.”

Would 1.21 jigowatts cover it?

anonymous coward#distance-computation circuits, in addition to the flux capacitor et. al. This gets less and less plausible when one imagines how Doc Brown could have constructed a second time-ship out of a 19th century train–as is revealed in the closing scene of BTTF 3. After-all, the retroengineering required in fashioning a flux capacitor from pre-period goods would be hard enough without having to fabricate silicon microprocessors as well.

Ah, but you forget Grassphopper, Doc Brown had the *original* flux capacitator AND indeed, the entire Delorean to work with, as long as he got it buried in the old mine, prior to Doc II and Marty I finding it! So the locomotive time vehicle could have easily been constructed, assuming that Doc waited (or worked in anticipation of) “valve” style vac tubes in the very early 20th century.

And not to overthink your overthinking on this one–but how exactly did you arrive at the conclusion that temporal translocation somehow deprived Einstein and the Delorean of their kinetic energy they already possessed, just like we do? Or to be less kind, how do YOU get any one minute into the future, without being dumped off into space, now out of sync with the Earth?

Michael#Along these lines, I’ve often pondered on using this phenomenon in the other direction. Forget time travel. Think SPACE travel. The fastest way to travel is to somehow phase yourself out of contact with the physical universe while staying anchored to the same absolute point in space. The longer you stay phased out, the further you went when you phase back in. No warp drive or wormholes required.

It seems damn hard to control this in a useful way, so you don’t wind up in the vast emptiness between nothing and nowhere, or inside a star, or the molten core of a planet, but you could cover immense distances in little time at all. Especially when you factor in all the big mindfuck variables you ignored in your article.

I’m glad I’m not the only one who thinks way too damn much.

I know where I can steal a Delorian. Can anybody help me figure out the rest of the equation?

graham#ok, i have to get zameckis’ back here. i think that he was much closer than you all think; the only problems are the rotation of the earth, and the exact time interval that is traveled.

first, the issue of motion around the sun is moot. in relativity, gravity bends space-time. the orbit of the earth’s center of mass is along a straight line through a curved space-time. since motion is relative, we can assume that the delorean’s frame of reference is stationary, so if it disappears and doesn’t move, it will end up in the analagous spot relative to the center of the earth traveling around the sun. this also deals with any movement of the galaxy, or galaxy cluster, or whatever; these are all linear motions through space-time that is bent by gravity. as long as the earth isn’t pushed by an electromagnetic field, or the strong or weak forces, we should be fine.

there is a problem, though, with the earth’s rotation. if the delorean were orbiting the earth, things would be fine, it would continue along that orbit while it was vanished (that is just like the issue of orbiting around the sun). however, it is not orbiting the earth, but is sitting on the surface and rotating. sitting there, it is being constantly accelerated away from its “stationary” trajectory through bent space-time, which would have it fall towards the center of the earth. so the most likely event would be that the delorean would disappear and re-appear embedded in molten rock. however, if the amount of time traveled were an integer multiple of the amount of time it took for one eliptical orbit around the center of the earth, it would end up back at the surface…

…but not at the same spot! the earth would have turned in the meantime, and the initial velocity of the delorean (the speed of the earth’s surface’s rotation +/- 88mph in some direction) would affect the shape of the parabola, but the extremity of that parabola, which returns the delorean to the surface of the earth (friction has no effect since it has disappeared), would put it at a different spot along the same latitude of the surface. now, for the longer trips, i am willing to suspend disbelief and believe that an interval of time was calculated that was evenly divisible by both the time taken by the parabolic journey around the center of the earth AND the 24-hour interval needed to return the appropriate part of the earth’s surface to the appropriate position. but for the short 1 minute trip (again, i’ll suspend disbelief and assume that “1 minute” means “the precise interval of time required for one parabolic trip around the center of the earth”), it would end up, as sheq points out, 23km out-of-place.

Peter B. Juul#You are all making this far too complicated.

There is no such thing as a universal reference point. Every observer is in his observation the center of the observed universe. That is what relativity is, on the most basic level.

All movement is relative to a reference point, and the most sensible reference point is the observer, ie. the center of the universe.

Einstein, McFly et al thus stays at the center of the universe while moving through time. And since the earth was only moving under them at 88 mph, when they “jump”, it still is when they “land”.

shechner#Oooh – Graham was SO close. But sadly, Earth’s orbit is not a line in space-time, but a helix.

SO, if we convert the entire system into polar coordinates, we can describe the orbit as a sine function in space-time, lending simplicity to the process, but not enabling us to jump the giant physical space- gap in space-time.

Also, Earth is not being pulled merely by the gravity of the sun, but by all other objects in the universe.

Still, good try!

Dave#Excellent article. Mistake in Part The First though, the earth rotates 360 degrees in approx 23h56m04s. In 24 hours, it’s rotated a little more, so that the sun is in the same place in the sky, but, as you also noted, the planet itself has moved. See “Sidereal time” on WikiP.

john#why couldn’t math be this much fun in school

Aaron#“any object passing through it would be exposed temperatures approaching absolute zero and would hence instantly become incomparably cold.”

DUN DUN DUNNN. This could explain something…

Good read.

Newt#It’s way worse than that.

According to wikipedia the solar system is orbiting the galactic center at a speed of 220 km/s, or thereabouts. And that’s not even the worst problem: the Milky Way itself is moving at about 552 km/s relative to the cosmic background radiation rest frame.

Rodney#I think you are forgetting one thing. The Delorean is always in contact with the earth, bound by gravity, so of course the car stays in the same physical location on earth. The time machine works relative to it’s place on the earth. Your theory only applies if the car leaves earth’s gravity, which it does not, it appears in the new time instantaneously as it leaves the old.

cjk#The fact that the car IS COLD after the dog’s timejump just gives a hint that a wormhole might indeed be part of the solution. (However, it does not seem cold the other times…)

DU#I don’t even want to go near the calculation for the 1985-1955 jump…Actually, this jump is easier to explain. If you remember, Doc Brown is the one who typed the 1955 date in (it was the day he first conceived of the flux capacitor, IIRC). Doc Brown being such a genius, he could well have already figured out that tonight (in 1985) the earth would be in the same absolute location that it was in 1955 and therefore any jump in time would put it in the same spot in space.

The same goes for 1885. He doesn’t need a microprocessor to do the work as long as he does it on paper first for the single case he’s interested in. Then jump to that time/location and install a computer….

Matt#See, my problem with BttF wasn’t with the relative position (even in the above example we assume relativity with the sun) because everything is relative and maybe they can set any point of reference they like and travel in the temporal dimension whilst maintaining that x,y,z reference point relative to whatever arbitrary point they choose. Rather, my problem was with the “it runs on steam” explanation for the train. I mean, a train-sized steam engine is generating 1.21GW of electricity?

Peter B. Juul#@matt: Plutonium steam, of course. Come on!

Steve McC#I used to worry about this with poor ol’ Rod Taylor. Based on his subjective perceptions from within the operating ‘classic’ Time Machine, I used to picture him watching helplessly as the earth spun away from him. He survived proximity lava flows and hundreds of millennia of encasement in solid rock in the movie, so presumably the cold, radiation and vacuum of spce would be no problem as long as he kept moving. The trouble then is, even to bring the Earth ‘back’ to exactly where you left so you weren’t above or below it but rather precisely on its surface as you had been positioned, would require stopping and reversing. Uh-oh…… and not being a college-level science student I have no idea whether his ‘H.George Wells’ character had enough astronomical science available in the day to possibly account for this in advance. Based on the finsihed movie therefore, he either DID, somehow, or else the guys above who lean towards an inherent spatial ‘lock-in’ for the time-travelling object are on the right track….

jin#don’t know if anyone brought this up yet but what about the whole “masters of space and time” effect where objects traveling back in time are actually LARGER? because the universe (AND EVERYTHING IN IT) is expanding.

so if you dislocate from your temporal point of origin, your relative size at your destination time is going to be severely out of whack.

jin

Steve#This is a dumb article. Everybody and their mother knows that the Flux Capacitor compensates for all this.

David Ermold#I love you for this.

jpkotta#An interesting interpretation of special relativity is that everything is moving with constant velocity. Not the regular spatial velocity that we’re used to, but 4-velocity. Time dilates for fast objects because they appear to be moving more in the space direction than the time direction of spacetime. Stationary objects don’t have time dilation because they move only in the time direction. No, I’m not going anywhere with this, just thought I’d share.

Joe Nahhas#Not only Einstein is worng but Also the 100,000 space-time physicists are wrong. These are real published cases where Einstein and all of physics and physicists can not solve

1-DI Her Binary stars motion

2- AS Cam Binary stars motion

3-V1143cgyni Binary stars motion

4-V514cgyni Binary stars motion

———————————

Thousands others cases Ignored by Physicists and Media where Einstein’s space-time is all bad physics

Einstein and all others are nothing but Naive believing in space-time continuum

It is time to uncover the greatest liars and they are Physicists

Einstein’s space-time physics in a failure like the Large hadronic collider project that burned itself

Joe Nahhas#For four hundred years physicists missed Kepler’s time dependent equation that expalin’s “relativistic” effects

as scientific “errors” not accounted for by Astronomers who for four hundred years still can not read a telescope

I challenge all physicists that space-time is nothing but an error and Einstein’s work is just mathematical garbage courtesy of handmen in white robes conducting “junk” experimentation and scientifically Ignorant media to expose the greatest Mistake that outdone W Bush errors

mrG#is there any reason to suppose that a time-jumping rift would NOT be geosynchronous? The quantum foam which we now know continually constitutes, de-constitutes and re-constitutes our local physical reality has no problem following along with geographical coordinates, so why then could a larger (200 year) foam rift not ALSO follow its geography?

AP#The approximation of the position vector is okay, for the Doc’s reference frame; he travels that distance in one minute (as the earth rotates). But what about Einstein’s position, and where is he during that minute?

If we are going to obey the laws of physics, and not violate the the relativistic universal speed limit we observe the following:

Einstien experience’s Doc’s minute as fraction of a second

Where the Doc travels ~1500 km, Einstein travels exactly 0km and never requires a velocity >88mph

This depends on the flux capacitors ability to distort space and time. At 88mph Einstein is “out of phase”, and the FC causes a gravitational / quantum mechanical distortion of space-time. Essentially bending dimensional space into a fold such that the two position vectors that were 1500km apart, are now connected in Einstein’s phase.

Therefore Einstien never has to exceed the speed of light, relativity is observered, proportionality of space and time vector’s is preserved, and BTTF is not ruined!

Lubricus#Surprisingly one very plausible way of time travel is actually already invented by the worlds most famous death metal band Dethklok. With the method it is no problem with different space frames and it’s possible to travel one hour into the future in one hour with the only risk of sufocation.

Lubricus#Damn when the earth rotates the angles also gets fucked up so the delorian needs to be rotated and the angular moment needs to be changed if no major crashes should occur!

What is absolute and what is relative, position, speed, acceleration, angles, rotation ???

kraken#I could be wrong here, as I’m not a physicist or mathematician, but all this calculation overlooks one important aspect:

Momentum.

I can’t speak for everyone, but I certainly do not feel any g-force while traveling at the rate of revolution and rotation of the Earth axis/around the sun. I can only assume that the momentum propelling us all and which is invisible to us would also envelope Einstein as he travels forward in time, thus he would appear in the same location and all that troublesome calculation just went to waste.

ShavenLunatic#I’m going to happily assume the FC takes all of this into account and teleports the vehicle to the required position… please don’t break BTTF any more :)

Erika#The trouble with all of this is, no matter where you were when you appeared in another time, you would be trying to occupy space that already had something in it. Unless you were in a perfect vacuum, and there is no perfect vacuum big enough for a human, not even in the space between galaxies.

This is a problem with teleportation, too. If you materialize where there’s already air, does the air just disappear? Or do you just disappear? Either would violate conservation of matter.

The only time travel or teleportation stories that work at all are the ones where you pass through some kind of portal, allowing you to displace matter as you enter your destination. So I have to agree with the wormhole idea.

nixo#I had a similar, though far less complex idea a number of years ago: Let’s say you’re in a field, and there is a brick hovering in the air in front of you, and a man standing near you says “In 10 seconds the brick will stop moving 0 mph relative to the ground here, but start moving 0mph relative to the center of the universe.”

Which way do you run?

Your best bet (if you live in the US) is south, of course you have less than 10 seconds to figure out which way south is.

MZ#So has anybody taken a stab at calculating the spatial displacement of Hill Valley when the DeLorean arrives in 1955?

Alex C#I thought of this problem long ago too! My personal scifi answer to this time-travel conundrum (similar to @Rodney and @hoof) was that passage along the time axis of spacetime would follow the curvature of space which includes effects of gravity (and, perhaps in a stretch, rotational inertia and angular momentum).

(OTOH, when traveling backwards in time, wouldn’t gravity become a repellent force?)

The real reason, of course, that time travel is impractical is that it hasn’t happened yet (where “yet” includes all future time too). You really couldn’t expect every single potential future time traveller to be well-behaved and not reveal himself to people in our time, could you? Reductio ad nauseam!

Eric#I think you have to assume that in any time travel scenario, the converservation of momentum is required.

Thus as with relativity and all other space-time constructs, we must consider the affects of the loss of mass in the future and the gain of mass in the past, for both frames.

I believe if you take this into consideration, you will realize that all calculations are useless until we measure the mass of the universe.

Joe Nahhas#Kepler (demolish) Vs Einstein’s

Ending Einstein’s space jail of time in 2009 that led to fraud Symbol E=mc²

Areal velocity is constant: r² θ’ =h Kepler’s Law

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity

r² θ’= h = S² w’

S = r exp (ỉ wt); h = [r² Exp (2iwt)] w’=r²θ’

w’ = (θ’) exp [-2(i wt)]

w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]

w’ = w'(x) + ỉ w'(y) ; w'(x) = (h/r²) [ 1- 2sine² (wt)]

Δ w’= w'(x) – (h/r²) = – 2(h/r²) sine² (wt) = – 2(h/r²) (v/c) ² v/c=sine wt

(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²

Δ w’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second

{x [180/π;degrees]x[100years=36526days;century]x[3600;seconds in degree]

Δ w” = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century

This Kepler’s Equation solves all the problems Einstein and all physicists could not solve

DI Her Binary starts systems

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

Advance of Perihelion of mercury.

G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg

ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec

Calculations yields:

v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552

Δ w”= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century

Conclusions: The 43″ seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler’s equation better than all of Published papers of Einstein. Kepler’s Equation can solve Einstein’s nemesis DI Her Binary stars motion and all the other dozens of stars motions posted for past 40 years on NASA website SAO/NASA as unsolved by any physics

Anyone dare to prove me wrong?

Joe Nahhas#E=mc²/2

2009 is the end of Einstein’s space-jail of time and Fraud symbol E=mc²

[email protected]

Time is not a structure like space to allow space-to time-back to space jumping claimed by Physicists regardless of what physicists have to say about it because Physics is a business and not necessarily science or scientific and like every business it comes with fraud and fraud is Einstein’s space-time (x, y, z, it) continuum that led to fraud symbol E=mc² and yes I am saying that 109 years of Nobel prize winners physics and physicists are all wrong and space-time physics is based on scientific fraud. When “results” expected and “No” discovery, Physicists rigged Physics for grant money since the start of the industrial revolution. Physics today is at least 51 % fraud!

r ——————>>Exp (ì w t) ———->> S=r Exp (ì wt) Nahhas’ Equation

Orbit——–>> Orbit light sensing——>> Visual Orbit; Exp = Exponential

Particle —->> light sensing of moving objects———— >> Wave

Newton———>>light sensing———->> Quantum

Quantum = Newton x Visual Effects

Quantum – Newton = Relativistic = Optical Illusions

E (Energy by definition) = mv²/2 = mc²/2; if v = c

m = mass; v= speed; c= light speed; w= angular velocity; t= time

S = r Exp (ì w t) = r [cos (wt) + ì sin (wt)] Visual effects

P = visual velocity = change of visual location

P = d S/d t = v Exp (ì w t) + ì w r Exp (ì w t)

= (v + ì w r) Exp (ì w t) = v (1 + ì) Exp (ì w t) = visual speed; v = wr

E (visual energy= what you see in lab) = m p²/2; replace v by p in E = mv²/2

= m p²/2 = m v²/2 (1 + ì) ² Exp (2ì wt)

= mv²/2 (2ì) [cosine (2wt) + ì sine (2wt)]

=ì mv² [1 - 2 sine² (wt) + 2 ì sine (wt) cosine (wt)];v = speed; c = light speed

wt = π/2

E (visual) = ìmv² (1 – 2 + 0)

E (visual) = -ì mc² ≡ mc² (absolute value;-ì = negative complex unit) If v = c

w t = π/4

E (visual) = imv² [1-1 +ỉ] =-mc²; v = c

wt =-π/4+ỉln2/2; 2ỉ wt=-ỉπ/2 – ln2

Exp (2i wt) = Exp [-ỉπ/2] Exp [ln(1/2)]=[-ỉ (1/2)]

E (visual) = imv² (-ỉ/2) =1/2mc² v = c

Conclusion: E = mc² is the visual Illusion of E = mc²/2 [email protected]. All rights reserved.

PS: In case of E=mc² claims to be rest energy claims then

E=1/2m (m v + m’ r) ² = (1/2m) (m’ r) ²; v = 0

E = (1/2m) (mc) ²; m’ r =mc

E=mc²/2

Alexander Nahhas#V1143Cgyni Binary Stars Apsidal motion Puzzle solution

The motion puzzle that Einstein MIT Harvard Cal-tech NASA and all others could not solve.

Introduction: For 350 years Physicists Astronomers and Mathematicians missed Kepler’s time dependent equation that changed Newton’s equation into a time dependent Newton’s equation and together these two equations combine classical mechanics and quantum mechanics into one mechanics explains “relativistic” effects as the difference between time dependent measurements and time independent measurements of moving objects and solve all motion in all of Mechanics posted on Smithsonian NASA website SAO/NASA that Einstein and all 100,000 space-time “physicists” could not solve by space-time physics or any published physics.

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location

r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location:

P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

= change of location + change of mass

= m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

F = d P/d t = d²S/dt² = Total force

= m(d²r/dt²) +2(dm/dt)(d r/d t) + (d²m/dt²)r

= mγ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

Proof:

r = r [cosθ î + sinθĴ] = r r (1); r (1) = cosθ î + sinθ Ĵ

v = d r/d t = r’ r (1) + r d[r (1)]/d t = r’ r (1) + r θ’[- sinθ î + cos θĴ] = r’ r (1) + r θ’ θ (1)

θ (1) = -sinθ î +cosθ Ĵ; r(1) = cosθî + sinθĴ

d [θ (1)]/d t= θ’ [- cosθî – sinθĴ= – θ’ r (1)

d [r (1)]/d t = θ’ [ -sinθ'î + cosθ]Ĵ = θ’ θ(1)

γ = d [r'r(1) + r θ' θ (1)] /d t = r” r(1) + r’ d[r(1)]/d t + r’ θ’ r(1) + r θ” r(1) +r θ’ d[θ(1)]/d t

γ = (r” – rθ’²) r(1) + (2r’θ’ + r θ”) θ(1)

F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m’[r'r(1) + rθ'θ(1)] + (m”r) r(1)

= [d²(mr)/dt² - (mr)θ'²]r(1) + (1/mr)[d(m²r²θ')/dt]θ(1) = [-GmM/r²]r(1)

d²(mr)/dt² – (mr)θ’² = -GmM/r² Newton’s Gravitational Equation (1)

d(m²r²θ’)/dt = 0 Central force law (2)

(2) : d(m²r²θ’)/d t = 0 m²r²θ’ = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ'(θ, t)]

= [m²(θ,t)][r²(θ,t)][θ'(θ,t)]

= [m²(θ,0)][r²(θ,0)][θ'(θ,0)]

= [m²(θ,0)]h(θ,0);h(θ,0)=[r²(θ,0)][θ'(θ,0)]

= H (0, 0) = m² (0, 0) h (0, 0)

= m² (0, 0) r² (0, 0) θ'(0, 0)

m = m (θ, 0) φ (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential

φ (0, t) = Exp [ λ (m) + ỉ ω (m)]t

r = r(θ,0) ψ(0, t) = r(θ,0) Exp [λ(r) + ì ω(r)]t

ψ(0, t) = Exp [λ(r) + ỉ ω (r)]t

θ'(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} ——I

Kepler’s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein’s space-jail of time

θ'(0,t) = θ'(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}

(1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²

d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)

Let m r =1/u

d (m r)/d t = -u’/u² = -(1/u²)(θ’)d u/d θ = (- θ’/u²)d u/d θ = -H d u/d θ

d²(m r)/dt² = -Hθ’d²u/dθ² = – Hu²[d²u/dθ²]

-Hu² [d²u/dθ²] -(1/u)(Hu²)² = -Gm³(θ,0)φ³(0,t)Mu²

[d²u/ dθ²] + u = Gm³(θ,0)φ³(0,t)M/H²

t = 0; φ³ (0, 0) = 1

u = Gm³(θ,0)M/H² + Acosθ =Gm(θ,0)M(θ,0)/h²(θ,0)

mr = 1/u = 1/[Gm(θ,0)M(θ,0)/h(θ,0) + Acosθ]

= [h²/Gm(θ,0)M(θ,0)]/{1 + [Ah²/Gm(θ,0)M(θ,0)][cosθ]}

= [h²/Gm(θ,0)M(θ,0)]/(1 + εcosθ)

mr = [a(1-ε²)/(1+εcosθ)]m(θ,0)

r(θ,0) = [a(1-ε²)/(1+εcosθ)] m r = m(θ, t) r(θ, t)

= m(θ,0)φ(0,t)r(θ,0)ψ(0,t)

r(θ,t) = [a(1-ε²)/(1+εcosθ)]{Exp[λ(r)+ω(r)]t} Newton’s time dependent Equation ——–II

If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then

θ'(0,t) = θ'(0,0) Exp{-2ì[ω(m) + ω(r)]t}

r(θ, t) = r(θ,0) r(0,t) = [a(1-ε²)/(1+εcosθ)] Exp[i ω (r)t]

m = m(θ,0) Exp[i ω(m)t] = m(0,0) Exp [ỉ ω(m) t] ; m(0,0)

θ'(0,t) = θ'(0, 0) Exp {-2ì[ω(m) + ω(r)]t}

θ'(0,0)=h(0,0)/r²(0,0)=2πab/Ta²(1-ε)²

= 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ'(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²

θ'(0,t) = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t

θ'(0,t) = {2π[√(1-ε²)]/(1-ε)²}{cos 2[ω(m) + ω(r)]t – ỉ sin 2[ω(m) + ω(r)]t}

θ'(0,t) = θ'(0,0) {1- 2sin² [ω(m) + ω(r)]t – ỉ 2isin [ω(m) + ω(r)]t cos [ω(m) + ω(r)]t}

θ'(0,t) = θ'(0,0){1 – 2[sin ω(m)t cos ω(r)t + cos ω(m) sin ω(r) t]²}

– 2ỉ θ'(0, 0) sin [ω (m) + ω(r)] t cos [ω (m) + ω(r)] t

Δ θ (0, t) = Real Δ θ (0, t) + Imaginary Δ θ (0.t)

Real Δ θ (0, t) = θ'(0, 0) {1 – 2[sin ω (m) t cos ω(r) t + cos ω (m)t sin ω(r)t]²}

W(ob) = Real Δ θ (0, t) – θ'(0, 0) = – 2 θ'(0, 0){(v°/c)√ [1-(v*/c) ²] + (v*/c)√ [1- (v°/c) ²]}²

v ° = spin velocity; v* = orbital velocity; v°/c = sin ω (m)t; v*/c = cos ω (r) t

v°/c << 1; (v°/c)² ≈ 0; v*/c << 1; (v*/c)² ≈ 0

W (ob) = – 2[2π √ (1-ε²)/T (1-ε) ²] [(v° + v*)/c] ²

W (ob) = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² radians

W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² degrees; Multiplication by 180/π

W° (ob) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

W” (ob) = (-720x26526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² seconds /100 years

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)

v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈ 0; m<<M

Application 1: Advance of Perihelion of mercury.

G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg; ε = 0.206; T=88days

c = 299792.458 km/sec; a = 58.2km/sec; 1-ε²/4 = 0.989391

ρ (m) = 0.696×10^9m; ρ(m)=2.44×10^6m; T(sun) = 25days

v° (M) = 2km/sec ; v° = 2meters/sec

v *= v(m) = √ [GM/a (1-ε²/4)]; v(M) = √[Gm²/(m + M)a(1-ε²)] ≈ 0

v°(m) = 2m/sec (Mercury) v°(M)= 2km/sec(sun)

Calculations yields: v = v* + v° =48.14km/sec (mercury); [√ (1- ε²)] (1-ε) ² = 1.552

W” (ob) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²

W” (ob) = (-720x36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century

V1143Cgyni Apsidal Motion Solution

W° (ob) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

v° = -v°(m) + v°(M)

v* = 2v(cm) + σ

v°(m) = spin velocity of primary

v°(M) = spin velocity of secondary

v(cm) = [m v(m) + M v(M)]/(m + M) center of mass velocity

σ = √ {{[v(m) - v(cm)]² + [v(M) - v(cm)]²}/2} = standard deviation

W° = 3.36°/century as reported in many articles

peter Garvin#the solution is obvious. the delorean always stays at the same point in space no matter where or when it drives to. the delorean moves the universe around it. solved

peter Garvin#and i think the theory of “if we ever do figure out time travel, why hasn’t there been any time travelers here?”

im not saying that there are but if there are time travelers here, what do you think would happen if they said they came from the year 3000? they would be locked away in an insane asylum. also, if they said anything about them being time travelers, it could very likely cause a major paradox.

also in bttf part 2, wouldnt that be a paradox? the alternate 1985 where doc is in an insane asylum. he wouldnt have built the time machine.

nick#@Alexander Nahhas or Joe Nahhas

What the bloody hell are you going on about? Quack.

c0nzo#what would happen if, on the time numeric pad, you put in like the 29th of febuary?

Dreamcatcher#Forget all this time-travel nonsense. What about Mr. Fusion??

Gab#http://www.youtube.com/watch?v=HDSsKdaW8ho&eurl=http%3A%2F%2Fnerdfighters.ning.com%2Fprofiles%2Fblogs%2Ftime-travel&feature=player_embedded

The “Evil Baby Orphanage” is an idea this guy and his brother have about going back in time, taking babies that grow up to do a lot of bad things to society to some facility, and raising them there instead of going back and actually killing them/their parents to stop whatever it is they end up doing as adults. The relative time travel stuff starts at about 1:15.