Outrunning Light; how quickly will light catch up with you

[CarsonAdams]

Outrunning Light; how quickly will light "catch up with you"

Thank you all in advance, this is my first official thread but certainly not my first time eyeing the discussions of these forums. Hopefully this enlightens us all.

Person A holds a flashlight
Person B rides a jet capable of near light speeds with instantaneous acceleration (or is alternatively flying by already at near light speed if you prefer not to swallow instantaneous acceleration) and is 100 meters from Person A.
Person C stands in what we will consider the "initial" reference frame as an observer to measure distances.

If person A holds a flashlight which will emit a single photon the moment an electric current from a circuit is received, and Person B will accelerate instantaneously to .99c the moment it receives an electric current from the same circuit with an origin current provider point equidistant (at the 50 meter point) from Person A and Person B, how far can Person B travel before the photon makes contact with him?

Understanding the relativistic effects of time dilation and velocities, from Person B's perspective, the photon will reach him, I conclude, in 3.3x10^-7 seconds (100 meters/the speed of light C) whether the jet begins moving or not, as the light will always travel at C and the jet can therefore always be treated as stationary to the light, only giving it 100 meters of displacement to travel.

What I don't understand is that surely the photon will not reach the jet instantaneously, which would imply that the jet has time to travel some distance with respect to the initial reference from as measured by Person C before being "caught" by the photon, just as if some Car A were traveling at 15 m/s and some Car B were traveling at 10 m/s, Car B would be able to travel some distance with respect to the road before being caught by Car A.

Due to light's non-relativitic speed, however, I'm confused as to how far, or perhaps how long it would take for the photon to reach Person B with respect to the initial reference frame of Person C. Will the time it takes for the photon to reach Person B with respect to Person C (denoted by Tc) differ from the time it takes for the photon to reach Person B with respect to Person B's reference far (Tb) by the relativistic constant for time dilation? As in Tc=γTb? And subsequently distance traveled can be calculated by velocity*time=distance travelled=γ(100m/C)(.99C)? or Would that then, to observer C, look as if light is not traveling at the speed of light?

(This question came from the research I've been conducting to discover the various escape velocities needed to outrun a nuclear blast if the "runner" begins accelerating as soon as he sees the bomb, not when the light from the explosion is seen, but the gamma and x rays produced will be so potent and intense at small distances away and will be virtually impossible to outrun at lower than relativistically relevant speeds that anyone within a certain range will receive lethal radiation doses. If, however, the "runner" could simply "stay ahead" of light for a few hundred or preferably over a thousand meters, you would, obviously, eventually lose to pursuant gamma rays, but would be past the lethal intensity zone as photons that had not been angled directly at you would have gone wildly off course from your trajectory by then)

 

[ghwellsjr]

I doubt that anybody can figure out what you are asking. You've got too many reference frames. You should have just one. Say where everyone and everything is at Coordinate Time zero. Then describe the speed of everyone and everything as a function of the Coordinate Time. Then ask your questions. After you figure out everything, you can transform the coordinates of each significant event into any other Inertial Reference Frame (IRF) moving with respect to the original one. Please don't think that every observer is in a different IRF. Everyone and everything is in every IRF but you need to deal with one at a time.

So can you please describe everything in terms of just one IRF?

 

[Dale]

Thank you all in advance, this is my first official thread but certainly not my first time eyeing the discussions of these forums. Hopefully this enlightens us all.

Hi CarsonAdams, welcome to PF!

Person A holds a flashlight
Person B rides a jet capable of near light speeds with instantaneous acceleration (or is alternatively flying by already at near light speed if you prefer not to swallow instantaneous acceleration) and is 100 meters from Person A.
Person C stands in what we will consider the "initial" reference frame as an observer to measure distances.

If person A holds a flashlight which will emit a single photon the moment an electric current from a circuit is received, and Person B will accelerate instantaneously to .99c the moment it receives an electric current from the same circuit with an origin current provider point equidistant (at the 50 meter point) from Person A and Person B, how far can Person B travel before the photon makes contact with him?

OK, .99 c is an annoying velocity and meters are a pain to work with, so I will just use d for 100 m and v for .99 c. You can plug in numbers later as you wish.

Understanding the relativistic effects of time dilation and velocities, from Person B's perspective, the photon will reach him, I conclude, in 3.3x10^-7 seconds (100 meters/the speed of light C) whether the jet begins moving or not, as the light will always travel at C and the jet can therefore always be treated as stationary to the light, only giving it 100 meters of displacement to travel.

What I don't understand is that surely the photon will not reach the jet instantaneously, which would imply that the jet has time to travel some distance with respect to the initial reference from as measured by Person C before being "caught" by the photon, just as if some Car A were traveling at 15 m/s and some Car B were traveling at 10 m/s, Car B would be able to travel some distance with respect to the road before being caught by Car A.

So in the initial reference frame you can write B's worldline as $B(t)=(ct,x_B(t))=(ct,vt+d)$ and you can write the photon's worldline as $P(t)=(ct,x_P(t))=(ct,ct)$. You can then solve for when the photon reaches B as follows:
$x_B(t_1)=x_P(t_1)$
$vt_1+d=ct_1$
$t_1=d/(c-v)$

Substituting that back in gives $x_P(t_1)=x_B(t_1)=cd/(c-v)$

Now, to get those coordinates in B's reference frame we simply do a Lorentz transformation.
$B'(t_1)=P'(t_1)= (ct'_1,x_B'(t'_1)) = (d \, \gamma,d\,\gamma)$
where $\gamma=(1-v^2/c^2)^{-1/2}$

 

[Nugatory]

We'll work the problem using a reference frame in which A is at rest because you described the initial conditions that way.

We'll measure distances in meters because that's how you described the problem, and we'll measure time in units of 1/3x10^-8 seconds so that the speed of light is 1.0: one meter per one unit of time. This avoids cluttering up the formulas with a bunch of c's.

We'll assume, as you wisely suggest, that B is already flying by at near lightspeed, so as not to have to mess with any accelerations.

And with those preliminaries out of the way...

Working in the frame in which A is at rest, we assign coordinates to the various events as follows:

t=-50, x=50: the electrical impulse is triggered.
t=0, x=0: the electrical impulse reaches observer A who launches the light signal.
t=0, x=100: the electrical impulse reaches observer B, who is moving at .99c and just happened to be there when the impulse arrived.
t=10,000, x=10,000: the light signal reaches observer B's spaceship. I calculated this from the fact that in A's frame, the light is moving at c and B's ship is moving at .99c with a 100 meter headstart. The light reaches B when $t=100+.99t$ and it's just algebra from there.

Now we want to know how these events appear to observers B and C. That's easy... Just use the Lorentz transforms to calculate the x and t values for each of these events as understood by observers moving relative to A. In particular (and I think this is what's behind your question) observer B will say he has a much smaller head start (Lorentz contraction of the distance between him and A) but will still calculate the light signal approaching him at speed c and catching up to him in much less than the 10,000 ticks that A observed (time dilation).

If in doubt... Draw a space-time diagram. There are plenty of examples in various recent threads.

 

[Simon Bridge]

The others have explained about the transformations ... I thought I may have noticed something else in there - just checking:

[...]Understanding the relativistic effects of time dilation and velocities, from Person B's perspective, the photon will reach him, I conclude, in 3.3x10^-7 seconds (100 meters/the speed of light C) whether the jet begins moving or not, as the light will always travel at C and the jet can therefore always be treated as stationary to the light, only giving it 100 meters of displacement to travel.

You mean the distance between where the light is emitted from and B is constant in B's reference frame?

The description you gave looks very confused and has lots of distracting details in it.
Rather than work through the confusions - I thought I'd try to give you clearer language to talk about your ideas, using an equivalent problem.

At t=0 on both their clocks, Alice (A) and Bob (B) pass each other with relative speed v.
If $a$ is Alice's position and $b$ is Bobs position, and I use subscripts A and B to represent their reference frames, then:

##[a]_A=0;\; _A=v[t]_A$... which is to say, according to Alice, she is stationary and Bob is moving forward at speed v;$[a]_B=-v[t]_B;\; _B=0##
... and according to Bob, he is stationary and Alice is moving backwards at speed v.

Also at t=0 on both their clocks, a light source which is visible to both of them, at position $[x]_A=-d$, changes color.
Your initial question amounts to asking what the time is on Bob's clock when he sees the change in color.

Your description looks like you meant $[d]_A = 100$m, but the calculation (quoted) looks like you meant $[d]_B=100$m.
In the rest of your post, it looks like you are confusing different reference frames.

By being careful about your descriptions, you will find the relationships with other reference frames easier to figure out.

Aside:
100m is a very short distance - light covers that distance in about 1/3,000,000s ... if you are more comfortable with actual numbers, you will find the math easier if you choose to measure distances in light-seconds. i.e. light takes 100s to cross 100ls and c = 1ls/s and v < 1ls/s. While you are making up examples, you may as well make up easy numbers eh?

 

[CarsonAdams]

I doubt that anybody can figure out what you are asking. You've got too many reference frames.

Duly noted, heartily agree after stepping back from the problem. Forgive my extraneous details, etc.

So in the initial reference frame you can write B's worldline as $B(t)=(ct,x_B(t))=(ct,vt+d)$ and you can write the photon's worldline as $P(t)=(ct,x_P(t))=(ct,ct)$. You can then solve for when the photon reaches B as follows:
$x_B(t_1)=x_P(t_1)$
$vt_1+d=ct_1$
$t_1=d/(c-v)$

Substituting that back in gives $x_P(t_1)=x_B(t_1)=cd/(c-v)$

Now, to get those coordinates in B's reference frame we simply do a Lorentz transformation.
$B'(t_1)=P'(t_1)= (ct'_1,x_B'(t'_1)) = (d \, \gamma,d\,\gamma)$
where $\gamma=(1-v^2/c^2)^{-1/2}$

An excellent interpretation of my ill-posed question and a wonderfully discreet mathematical explanation. I expected the displacement that B would attain in the IRF to be more complex than the differences between their velocities, generally given by
$d'_2=v_1d_2/(v_1-v_2)$
as per the classical calculation. The relativistic effects, I now see, are only experienced by B in his respective RF.
I suppose I had lost my way, thanks for bringing me back, extremely helpful and fast members.