Particles traveling back in time

[nitsuj]

Yes there is only one of you (classically), but its not the 3 dimensional object you are thinking of. Rather its a very complicated 4 dimensional object that CAN (in theory) come back and "interact" with itself (interact = intersect).

Weird, so the matter is duplicated. I can't for the life of me see this as conceivable. To your point the dimensions cannot be separated (4D), taking me from light path a in space and place me in a earlier place along that time path in space. Time cannot circumnavigate a length & vice versa.

I don't find 4D complicated and am not sure which 3D object you have me thinking of, or how the alternative (4D) differs. If you can see so clearly the 3D -4D distinction how can the absurdity of backward time travel even be considered. It can intersect at the same instant. interact, no. Go "before" the instant, no not even close.

There is only one of me no matter the metric, this is up to and including GR. Another way to say it there is only one "present moment" for any object in spacetime.

 

[WannabeNewton]

Any Lorentz boost can be decomposed into consecutive infinitesimal transformations belonging to the connected component of $O(3,1)$ which contains the identity transformation (the decomposition will start with the identity transformation). The time reversal transformation (which is non-orthochronous by definition) belongs to a connected component which does not contain the identity. Lorentz boosts are orthochronous transformations in the above sense i.e. they preserve the notion of "future-directed" or "past-directed" of vectors.

 

[PeterDonis]

if you boost a particle (-E, -p) by v (the particle is moving in the -v direction) you find that the resulting 4-momentum is (-m,0)!

Yes, and that indicates to me that the alternate solution in my previous post is the correct one: your picture was wrong. A particle with negative energy and negative x-momentum must be moving in the *positive* x-direction, not the negative x-direction. So *both* particles are moving in the positive x-direction from the spontaneous creation event--i.e., both worldlines slope up and to the *right* in your picture.

Basically if a particle has negative energy and you perform a Lorentz boost in 1 direction, it will speed up in that direction instead of slow down!

That's not quite right; what happens is that the meaning of spatial momentum is reversed for a particle with negative energy, as above.

Here's another way of seeing it: look at the 4-velocity of such a particle (which, with a constant rest mass, is just the normalized 4-momentum). The 4-velocity components in a given frame are $(dt / d\tau, dx / d\tau)$. For a particle with negative energy, $dt / d\tau$ is negative; and for a particle with negative x-momentum, $dx / d\tau$ is negative. But that means that $dx / dt = (dx / d\tau) / (dt / d\tau)$ is *positive*; which means, as I said above, that a negative energy particle with negative x-momentum is moving in the positive x-direction, just like a positive energy particle with positive x-momentum. So both worldlines overlap completely; that allows the two particles' energy and momentum to cancel each other out in all frames.

This is indeed somewhat bizarre, as you say, but I agree it's much less bizarre than violating the invariance of 4-momentum conservation.

 

[michael879]

Yes, and that indicates to me that the alternate solution in my previous post is the correct one: your picture was wrong. A particle with negative energy and negative x-momentum must be moving in the *positive* x-direction, not the negative x-direction. So *both* particles are moving in the positive x-direction from the spontaneous creation event--i.e., both worldlines slope up and to the *right* in your picture.

Ok I see what your saying, it makes perfect sense, but I can't get it to come out. If you're right can you explain where the flaw in my logic is?

left particle:
x' = (t',x')
u' = dx'/dτ = -γ(1, dx'/dt')
p' = mu' = -γm(1, dx'/dt')

right particle:
x = (t,x')
u = dx/dτ = γ(1, dx/dt)
p = mu = γm(1, dx/dt)

dx/dt = dx'/dt' = v

Therefore the 3-velocity of the left particle would be -v and it should be moving to the left...

*edit* Similarly: dx'/dt = dx'/dt' * dt'/dτ * dτ/dt = v * (-γ) * (1/γ) = -v

 

[PeterDonis]

can you explain where the flaw in my logic is?

I don't think there's any flaw in what you wrote up to your final paragraph (see below), except that your variable naming seems a little confusing; the variable "v" is usually used to mean dx/dt, not dx/dtau, but that's a detail. You're just agreeing with me that both particles have positive dx/dt. But that means both of their worldlines will slope up and to the right in a spacetime diagram; your diagram had the negative energy particle's worldline sloping up and to the left.

Therefore the 3-velocity of the left particle would be -γv and it should be moving to the left...

No, its 3-velocity is its dx/dt, not its dx/dtau. Both particles have a positive 3-velocity.

 

[michael879]

Sorry, you commented while I was in the middle of editing my post, please reread it? v is defined as dx/dt of the right particle

*edit* I just noticed I used v for both 4-velocity and 3-velocity. I replaced it with "u" for 4-velocity

 

[michael879]

No, its 3-velocity is its dx/dt, not its dx/dtau. Both particles have a positive 3-velocity.

The 3-velocity would be dx'/dt not dx'/dt', and since dt = -dt' you end up with a negative velocity..

 

[PAllen]

The 3-velocity would be dx'/dt not dx'/dt', and since dt = -dt' you end up with a negative velocity..

You don't use a separate reference frame for each particle. You can represent both particles in anyone reference frame, but it is nonsensical to combine statements from two different frames. Thus, Peter is right.

 

[PeterDonis]

The 3-velocity would be dx'/dt not dx'/dt', and since dt = -dt' you end up with a negative velocity..

I'm getting confused with your variables; I'm not sure what the primed and unprimed ones are supposed to mean. But I think dt and dt' are just the coordinate time differentials for the two particles; and those are both positive, because we're looking at both particles' worldlines going into the future, not the past. In other words, dt and dt' refer to two different differentials of the *same* coordinate time, t; they do not refer to differentials of two different time coordinates. (I see PAllen has made this point too.)

To make this clearer, let me go back to the two 4-momentum vectors, which are the fixed points we know we agree on. They are $(E, p)$ for the positive energy particle and $(- E, - p)$ for the negative energy particle. Factoring out the rest mass $m$, which is constant, we have the two 4-velocities $(\gamma, \gamma v)$ and $(- \gamma, - \gamma v)$. So far, so good.

But what do the 4-velocity components mean, physically? They are rates of change of the $t$ and $x$ coordinates with respect to *proper* time, right? So if we label the positive energy particle's coordinates as $(t_p, x_p)$, and the negative energy particle's coordinates as $(t_n, p_n)$, we have

$$
\frac{dt_p}{d\tau_p} = \gamma
$$
$$
\frac{dx_p}{d\tau_p} = \gamma v
$$

for the positive energy particle, and

$$
\frac{dt_n}{d\tau_n} = - \gamma
$$
$$
\frac{dx_n}{d\tau_n} = - \gamma v
$$

for the negative energy particle. These are easily integrated to obtain parametrizations of the two worldlines:

$$
(t_p, x_p) = \tau_p ( \gamma, \gamma v)
$$
$$
(t_n, x_n) = \tau_n ( - \gamma, - \gamma v)
$$

But we still have two $\tau$ variables; how are they related? We'll assume they are scaled the same, so the only question is the sign. Suppose the signs are the same. Then we have

$$
(t_p, x_p) = \tau ( \gamma, \gamma v)
$$
$$
(t_n, x_n) = \tau ( - \gamma, - \gamma v)
$$

If we assume that $\tau = 0$ is the spontaneous creation event, and that $\tau > 0$ describes the two worldlines after that event, then we have a problem: the two worldlines don't describe the "V" shape you drew. Instead, they describe two segments of an infinite single worldline from t = minus infinity to t = plus infinity, with the "spontaneous creation" event simply being whichever event on this single worldline we pick as the origin of $\tau$.

So now suppose the signs are opposite; then we have
$$
(t_p, x_p) = \tau ( \gamma, \gamma v)
$$
$$
(t_n, x_n) = \tau ( \gamma, \gamma v)
$$

In other words, the two worldlines both go into the future (increasing $t$) from the spontaneous creation event, but they overlap--they are identical, so they cancel each other out.

The upshot of all this is that the picture you drew can't be right, no matter which assumption we make about $\tau$. This "negative energy particle" business can't possibly describe a "V" shaped pair of worldlines; it can only describe either a single fully infinite worldline, or two half-infinite worldlines that overlap and cancel.

 

[michael879]

If we assume that $\tau = 0$ is the spontaneous creation event, and that $\tau > 0$ describes the two worldlines after that event, then we have a problem: the two worldlines don't describe the "V" shape you drew. Instead, they describe two segments of an infinite single worldline from t = minus infinity to t = plus infinity, with the "spontaneous creation" event simply being whichever event on this single worldline we pick as the origin of $\tau$.

In other words, the two worldlines both go into the future (increasing $t$) from the spontaneous creation event, but they overlap--they are identical, so they cancel each other out.

The upshot of all this is that the picture you drew can't be right, no matter which assumption we make about $\tau$. This "negative energy particle" business can't possibly describe a "V" shaped pair of worldlines; it can only describe either a single fully infinite worldline, or two half-infinite worldlines that overlap and cancel.

Again, I understand what you're saying and I really appreciate the detailed, well thought out response. However, why would the intersecting wordlines necessarily need to go from -∞ to ∞? Aren't all conservation laws obeyed if both particles simply cease to exist at the intersection point? This would be the time-reversal of my original question.

Now I understand that so far we've assumed the particles are non-interacting, and an annihilation of both would be a type of interaction. So what about instead of annihilation we allow the particles to have some type of elastic interaction with each other? At the "collision" point there would be an infinite number of possible outcomes that preserve both energy and momentum (which one would depend on the details of the interaction)! One of these possible outcomes would be both particle coming to rest, where they would overlap and effectively "annihilate" leaving space-time empty again! I don't see any law preventing this scenario...

 

[PeterDonis]

However, why would the intersecting wordlines necessarily need to go from -∞ to ∞?

I didn't assume they would; I derived that conclusion for the case where $\tau_p = \tau_n = \tau$. The only assumption I made, other than the intial 4-momentum vectors, was that each worldline is "half-infinite", which is what you postulated in your original scenario. The rest pops out of the math, no assumptions required.

Aren't all conservation laws obeyed if both particles simply cease to exist at the intersection point? This would be the time-reversal of my original question.

Sure, in my notation that would be the case where $\tau_p = - \tau_n = \tau$ and $\tau <= 0$ for both worldlines ($\tau = 0$ would be the "spontaneous destruction" event in this case). And in that case, we would have two overlapping worldlines coming from t = minus infinity and cancelling each other out. We still wouldn't have a "V" shaped pair of worldlines.

At the "collision" point there would be an infinite number of possible outcomes that preserve both energy and momentum (which one would depend on the details of the interaction)!

Sure, this just corresponds to the infinite number of possible choices for $v$ in the equations I wrote down. But the conclusions remain the same; what I wrote down is valid for any $-1 < v < 1$. (I put the choice in terms of $v$ rather than $\gamma$ because the sign of $v$ can be positive or negative, as I just showed, so the mapping from $v$ to $\gamma$ is not invertible--a given $v$ and $- v$ map to the same $\gamma$.)

Once again, the only assumptions I made were the initial 4-momentum vectors, which have to sum to zero net energy and momentum by the conservation laws, and that each particle's worldline was "half-infinite". Everything else follows from that. You appear to be looking for extra degrees of freedom in the problem that aren't actually there.

 

[michael879]

I didn't assume they would; I derived that conclusion for the case where $\tau_p = \tau_n = \tau$. The only assumption I made, other than the intial 4-momentum vectors, was that each worldline is "half-infinite", which is what you postulated in your original scenario. The rest pops out of the math, no assumptions required.

You couldn't have assumed that they were half-infinite because your conclusion was that they were both infinite!

 

[PeterDonis]

You couldn't have assumed that they were half-infinite because your conclusion was that they were both infinite!

No, that wasn't my conclusion. Look at the parametrizations I wrote down, carefully. For the case $\tau_p = \tau_n = \tau$, we have

$$
(t_p, x_p) = \tau ( \gamma, \gamma v)
$$
$$
(t_n, x_n) = \tau ( - \gamma, - \gamma v)
$$

where $\tau >= 0$ for both worldlines (and $\tau = 0$ is the point where they meet). So the positive energy particle's worldline starts at $(t, x) = (0, 0)$ and goes to t = plus infinity, and the negative energy particle's worldline starts at $(t, x) = (0, 0)$ and goes to t = minus infinity. Both are half-infinite.

(Of course this could equally well be interpreted as describing a single positive energy particle going from t = minus infinity to t = plus infinity. That was part of my point. But if you insist on calling the part with negative t the worldline of a "negative energy particle", then we have two "particles" each with a half-infinite worldline.)

 

[michael879]

ooo ok I think I realized my mistake. I was confusing proper time with the time coordinate (i.e. I was using the "lab time" coordinate as τ and two different time coordinates t and t'). Thank you, this answer makes far more sense and makes the concept of particles traveling backwards in time much more sensible. Like you said, to conserve energy and momentum the "spontaneous" creation of positive/negative particles must combine into a single worldline that is either (E,p) or (0,0). This essentially forbids the process, which means you can allow particles to travel back in time and remain consistent with both theory and experiment!

 

[remmeler]

How do you reconcile the problems of going backward in time

How can you reconcile the problems with time travel
Everyone know about killing you grandfather but how about -

Traveling 5 days back in time and for some reason not being able to travel forward except in the normal way of 1 day at a time.

There would be two of you existing in that time, but in 5 days your original self would travel back again and now there would be two of you who have traveled back and so forth.

The only way it could happen would be that you traveled back in time and ended up in that part of the multi universe where you traveled back in time.

Also there is the problem that no one has ever met someone that had information that would prove that they traveled back in time.

Just a little stupid thought I had, just to see if anyone has any thoughts on it.

 

[nitsuj]

The only way it could happen would be that you traveled back in time and ended up in that part of the multi universe where you traveled back in time.

Can you answer to yourself what that quote of your means?

 

[remmeler]

To Nitsuj - There is a theory that every alternative has happened in a unlimited amount of universes in a multi universe. I will have to look it up, but the theory is there are duplicates of everyone having taken every decision.

I don't say I subscribe to this theory but it is one of the theories out there that has some professional credence. I may not be describing it exactly. But if you are curious, I will look it up.

 

[remmeler]

To Nitsuj - I did not yet go back to my original source but I did find this which seems a close approximation of what I had seen described.

Many-worlds interpretation of quantum mechanics

Hugh Everett's many-worlds interpretation (MWI) is one of several mainstream interpretations of quantum mechanics. In brief, one aspect of quantum mechanics is that certain observations cannot be predicted absolutely. Instead, there is a range of possible observations, each with a different probability. According to the MWI, each of these possible observations corresponds to a different universe. Suppose a die is thrown that contains six sides and that the numeric result of the throw corresponds to a quantum mechanics observable. All six possible ways the die can fall correspond to six different universes. (More correctly, in MWI there is only a single universe but after the "split" into "many worlds" these cannot in general interact.) Tegmark argues that a level III multiverse does not contain more possibilities in the Hubble volume than a level I-II multiverse. In effect, all the different "worlds" created by "splits" in a level III multiverse with the same physical constants can be found in some Hubble volume in a level I multiverse. Tegmark writes that "The only difference between Level I and Level III is where your doppelgängers reside. In Level I they live elsewhere in good old three-dimensional space. In Level III they live on another quantum branch in infinite-dimensional Hilbert space." Similarly, all level II bubble universes with different physical constants can in effect be found as "worlds" created by "splits" at the moment of spontaneous symmetry breaking in a level III multiverse.

Related to the many-worlds idea are Richard Feynman's multiple histories interpretation and H. Dieter Zeh's many-minds interpretation