# 2 simple special relativity thought experiments

Tags:
1. Jun 25, 2015

### JustinM523

Hi. I'm learning about special relativity and want to see whether I understand it correctly. As such, I had 2 thought experiments that I answered, and I want to see whether these answers are correct. If not, could someone please tell me the correct answers? Thanks!

Thought Experiment 1: Super long straight-away with a start line and a big digital clock at the start line. Person1 (who had a running start) travels at a constant speed of half the speed of light and leaves the start line at "0" seconds. When the start clock hits 1 second (based on an observer standing at the start clock). a light beam (a photon) is fired straight ahead from the start line. How long does it take the light beam to reach Person 1?
1. From Person1's vantage point, he left when the start line said 0. He got 93,000 miles away and by that time his watch showed 1 second, but looking back at the clock at the start line, it appeared to show 0.5 seconds to him (from an "outside omniscient observer," this is objectively when the light beam left, and the start line actually showed 1 second). The light beam caught Person1 when his watch showed 2 seconds, but the start clock showed 1 second from his vantage point if he were to be looking back.
2. So I guess when we say "light approaches all constant-speed observers at 186,000 meters/second," in this case, we're saying that based on the time the observer sees on the clock from the start vs. end time. And the light did not "age" at all because it showed 1 second on the clock when it left and when it arrived from its vantage point.
3. From the vantage of a Person2 standing next to the start clock, I think the clock showed 3 seconds when it appeared the light beam caught Person1, which is 2 seconds after the light beam left.

Thought experiment 2: Let's say the universe is super big with, oddly enough, no movement of any celestial bodies, and you live on Planet 1 (far west part of the universe). You get shot at 99.9999% the speed of light toward Planet 2 (far east -- trillions of light years away). Ignore the fact that you would have a ridiculously large amount of mass and not actually live a normal life, and just assume you age like a normal person (someone who dies after ~80 years). Assume Planet 3 is 80 light years east of Planet 1. Do you die before you reach Planet 2?
1. My answer is yes, you die around the time that you pass Planet 3.

Last edited: Jun 25, 2015
2. Jun 25, 2015

### marcus

Hi Justin, your question seems more about understanding Special Relativity (SR) so you might get more helpful response in the SR forum. It is not about cosmology exactly. In cosmology there is a preferred frame and a preferred time (not the case in SR.) In cosmology space is expanding, distances between stationary observers increase over time. Distances can increase faster than the speed of light and without anybody moving thru their surrounding space, because that can happen with dynamic (i.e. changing) geometry. Again that is not the case with SR.

3. Jun 25, 2015

### JustinM523

Oh, yes, you are right. Sorry about that.

But by happenstance I've been of the opinion since learning about SR that there is a theoretically objective point of reference/time, so I'm glad you confirmed that.

4. Jun 25, 2015

### pervect

Staff Emeritus
I'm not sure why you think this. Is there a question here? I could try and give an answer, but I'm not sure what the question really is. It would also probably be better if we left out "outside omniscient observers", because it's rather likely that they don't obey the rules of special relativity.

You can't go at the speed of light, so the question as it stands asks us to assume something impossible, not a good start for asking a question. A question we can answer is "if we are shot at a high velocity approaching the speed of light, what does our watch read when you reach planet 2". The answer is it can read an arbitrarily small number, the formula is not distance / velocity, but rather distance / (velocity * gamma), where gamma = 1/sqrt(1 - v^2/c^2).

5. Jun 25, 2015

### JustinM523

Thanks for the comment.

I think I was sloppy in my phrasing, so I edited the question.

The question for the first one is, from various vantage-points, what would a wrist-watch show, and what would you see on the start clock, immediately before the light is "fired" and at the moment the light catches Person1?

For the second question, what I'm wrestling with is the idea that light does not age whatsoever. So, if someone were theoretically living life on a light beam that tried to cross the universe across trillions of light years, when that person would typically die at age 80 (and he passes Planet 3 after 80 light years), does he die before reaching the other side of the universe? (I'm not asking "does he age?" because I think that is too ambiguous and depends on who the observer is). At any rate, I understand he would have infinite mass if he were actually traveling at the speed of light (which I understand is not possible), so assume he is traveling very close to the speed of light.

6. Jun 25, 2015

### Janus

Staff Emeritus
From person 1's frame the light leaves the start line when his own clock read 1.15 sec and arrives when his clock reads 1.732 sec, at which time the start clock reads 1.5 sec and he will see the start clock read 1 sec. This also means that the time he sees on the start clock at the moment the light left is 0.577 sec, (start clock actually reads 1 sec)
Note that there is a difference between the time on the starting clock at any given moment according to person 1 (In Blue, and the time person one reads as the time through information carried by the light (In Green)

Using the same format (Red actual time on Person 1's clock, Blue, actual time on Start clock and Green the time the start clock sees on Person's ones clock):

According to the start clock when its clock reads 1 sec, it send the light signal. At that time, person 1's clock reads 0.866 sec and the the start clock would see it as reading 0.577 sec.

The light reaches person 1 when his clock reads 1.732 sec, and the start clock reads 2 sec and the start clock would see person 1's clock read 1.15 sec.

7. Jun 26, 2015

### Chalnoth

Huh? No, there is no objective point of reference/time. The foundational assumption of Special Relativity is that all observers in constant motion see identical laws of physics. In particular, they always measure the speed of light in a vacuum to be c. Things get a bit more complicated in General Relativity, but SR is good to get a lot of the basic concepts down in a simplified model.

You can pick one observer with one particular constant motion to be your "reference" observer. But another person could decide to pick a different observer, and there's no reason to think one is a better choice than the other. So usually the rest frame for a given calculation is chosen to make the calculation simple.

8. Jun 27, 2015

### Stephanus

Dear JustinM23, I'm learning SR, too (if not trying to understand), Can I ask here?
$V = 0.5$
$\gamma = \frac{1}{\sqrt{1-V^2}} = \frac{1}{\sqrt{1-0.25}} = 1.1547$
In Person 1's frame, it's Person 0 who travels. So person 0's clock is dilated. 1 second from person 0's clock is person 1's $1 sec * \gamma = 1.1547$ seconds.
So P1 sees P0 travels 0.5c away from him. Length is contracted/expanded, time is dilated/shortened, but speed remains the same. Is this true? And in 1.15 seconds (1 seconds wrt P0), P0 emits light to P1. P0 is 0.5774 ls away from P1. So the light will reach P1 at 1.1547 + 0.5774 = 1.7321 in P1 clock
But there's no way for P0 to know when the light will reach P1, is this right? Unless at the moment the light reaches P1, P1 sends a signal back to P0 and P0 calculates the time it first emits the light and the time P1's signal reaches P0 is 3 second.
I'm sorry. I'm lost here.In P0 view, when his clock reads 1 second, it emits a light to something (P1) that is moving away 0.5c. So the light will reach that thing in 1 second, right. LIke Zeno Paradox, Achilles catches up a turtle. When the turtle (P1) is 0.5ls away, Achilles (light from P0) travels to that distance. When Achilles reaches 0.5 ls, the turtle has moved forward 0.25 ls, when Achilles reaches 0.25ls again, the turtle has moved 0.125 ls, and on, and on until the light has reached 1 sec so it meets P1.
In P0's frame. The light will reach P1 at 1 ls away from P0.
In P1's frame. Something that is 0.5774 ls away from P1 emits a signal, while P1 is traveling 0.5c away. Of course we can't say that in P1 frame, P1 is moving away 0.c from the source light. Or we can? There's no Achilles/Turtle situation in P1 frame.The light will reach P1 in 0.5774 seconds, is this right? I think the answer is 1 second or 2 second if you count P0 emits a light when P0's clock already shows 1 seconds. Why 1.5? Can someone explain?
Of course if the moment P0 emits a light and P0 also sends a digital signal to P1 containing P0's clock. P1 will read the data in the signal 1 sec, you can also use a very big clock. But there's no way for P1 to know WHEN P0 emits the light, with a very big clock, is it? Unless P1 calculate speed and his clock (1.7321) and Lorentz contraction formula to deduce that P0 sends the signal at 1.1547 so P0's clock is 1 second. But 1 second is what the "Big Clock" shows when the light reaches P1, is this right?
Now, I'm lost here again.
Perhaps I should use the train simulation again.
Thank you very much Janus for introduce me the train simulation. I finally can devise Lorentz contraction from that simulation.
Okay...
Supposed P0 and P1 have friends.
Here is the diagram shown
...........A0...........B0...........P0...........C0...........D0
...........A1...........B1...........P1...........C1...........D1

The distance between P0, A0 and B0 is 0.5ls And A1, B1 and P1 is 0.5ls. A1, B1, P1 are in the same frame. So do P0, A0, and B0.
All their clocks are synchronized.

Frame 1 moves at 0.5c for 1 second. This is how the picture looks like for Frame 0.
Frame 1 is length contracted.

This is WHEN Frame 0 emits the light.
______________________________________|
..........A0...........B0...........P0|..........
C0...........D0
............................A1.......Z..B1.......P1.......C1.........D1

But in Frame 1, this is what Frame 1 sees
Frame 0 is length contracted.
This is the time P1 would have
__________________________________________________|
............A0.........B0.........P0.........C0..X........D0
.....................A1...........
B1...........P1|..........C1...........D1

Is this how the graph should look?

In Frame 0, does P0 emits light when P1 is at the same position as C0?
In Frame 1, does P0 emits light when B1 is at the same position is P0?
Is this why, if P1 would record the time P0 emits the light, P1 would read the clock in X? Hence 0.577. 0.5 * Gamma?
And if P0 reads Frame 1 clock when P0 starts emits the light (Z), it also read 0.577?
Shouldn't it 1.577? Because the events starts at T = 1, not T = 0.
There is a conclict between Frame 0 and Frame 1, is this the relative simultaneity?
For Frame 0. P0 meets B1 BEFORE C0 meets P0.
For Frame 1. P0 meets B1 AFTER C0 meets P0.
So the length contraction/time dilation is mutual, right?
For P0, 1 second is 0.866 second, or 1.15 second is 1 second.
But for P1, 1.15 second is 1 second, in an easier format
P0: 1.15 second is 1 second
P1: 1.15 second is 1 second

Yes. If we take Achille/Turtle situation, I'm sorry, I'm not used with SR. The light will reach P1 at 2 second. And for P1, it's 2 seconds / gamma = 1.7321 second
I don't know it all the calculations above are right. But thanks for any help.

9. Jun 27, 2015

### Stephanus

I'm sorry. There's one thing I forgot to calculate.
Red is in P1 frame.
Blue is in P0 frame.

Is this how to do it?
What is 1 second from an object who travels 0.5c?
So..
$1 = \sqrt{t^2 - (0.5t)^2}$
$1 = \sqrt{0.75t^2}$
$t = 1.1547$
So pink line is 1.1547, therefore the orange line from the equation y = 2x, must be 0.5774.

10. Jun 27, 2015

### Stephanus

Hi Justin, can I join the discussion?
From here on it's best to ask
How long does it take from P0 frame of reference the light beam to reach P1? 1 second. As I answered above, I calculate it with simple calculation.
The light, travels at c, is pursuing something that travels 0.5c at 0.5 sec * c away. So in mathematic
$ct = 0.5c + 0.5ct$
$t=0.5+0.5t$
$t = 1$
How long does it take from P1 frame of reference the light beam to reach P1?
This involves Lorentz contraction which I haven't mastered it yet. So I don't think I am capable to answer you. Janus, Marcus and some mentors above have answered it. But you have to add "from P1 frame of reference" in your question. That I'm sure.
This has been answered above. Remember 93,000 miles aways if from P0 frame, in P1 frame, it's different.
Of course no. Absolutely not odd
That would be very unlikely, but okay...
Planet 1 far west of the universe, Planet 2 far east of the universe. You mean both at the edge of the universe? It's true, if the universe is flat, or open. But in a closed universe, Planet 2 could be just west of Planet 1. Trillions of light years away? That's surely wrong. The "size" of the universe is "only" 93 billions ly. Open/Close/Flat is still being debated among scientist, but trillions ly is surely wrong.
Yes, you'll be right dead. Much less reaching Planet 3, You'll be smashed if you get shot at 99.99% of the speed of light without inertial dampener like in Star Trek . Okay..., say you'll survive the terrible acceleration, so using your number 99.9999% of the speed of light..., is that 4 9s after decimal point?
So, your time is dilated by using Lorentz Factor https://en.wikipedia.org/wiki/Length_contraction
$\gamma = \frac{1}{\sqrt{1-0.999999^2}}$ Is that 6 9s?
$\gamma = 707.707$, (calculator here). You'll only age $80/707.707/0.999999 = 0.11313717$ years. Almost 1 month.
What does the question about Planet 3, 80 years away from planet 1, have anything to do with Planet 2, trillion of ly away?
I'll try to do the math here.
If you want to reach Planet 2, say 93 billions ly away from you in you 80 years. So...
$\frac{(93 billions) * \sqrt{1-v^2}}{v} = 80$
$V = 0.9999999991$ calculator here. $V = 99.99999991\%c$
I'm struggling to understand SR, too. I think it's best if I exercise myselft with an "easy" question, see if I, too, understand it.

Last edited: Jun 27, 2015
11. Jun 27, 2015

### Janus

Staff Emeritus
Here are the S-T diagrams for the same scenario:

The left diagram is for the Person 1 frame, and the Right diagram is for the Start clock frame.

The yellow lines represent light carrying information from start clock to person 1 and vice versa. The small circles mark out time reading for each clock. When comparing the times on the clocks, the clock readings that are on the same horizontal line in a diagram are the simultaneous readings for those clocks for that frame.( for example, in the right image, or rest frame for the start clock, the start clock reads 1 sec at the same moment person 1's clock reads 0.866 sec. and at that moment the start clock is getting the light from person 1's clock when it read 0.577 sec, so that is the time he sees on person 1's clock)

One thing to note is that while the two frame might not agree on the simultaneous readings of each others clocks, they do agree on the clock readings when particular light signals leave and arrive at clocks. ( for instance both frames agree that the light leaving person 1's clock when it reads 0.577 sec arrives at the start clock when it reads 1sec).

12. Jul 5, 2015

### JustinM523

First, thank you to all who responded. This is very enlightening, and I certainly appreciate your help. Sorry for the lag on my side between replies.

This seems to be my source of confusion. Assume the start clock is a digital clock. Wouldn't the photons/light from the clock reach P1 with the same speed and in the same amount of time that the light hit P1? If the light leaves at the moment the clock hits 1 second (according to someone standing right next to the clock), I am confused as to why, from P1's perspective, the clock would not simply show "1" at the same moment that the light hits him (given that the light from the 1 in the clock should be leaving at the exact moment and with the exact speed as the separate photon of light).

The question was really more, "Planet X is 1,000 light years away. Humans live for 80 years. Bob is 1 year old. If Bob goes very very very near the speed of light, would he reach Planet X before he dies (ignoring things like acceleration, not having enough food, potentially going crazy, potentially hitting a black hole along the way, etc.)" The basic issue was whether traveling a light year at the speed of light causes 0 aging or 1 year of aging for every light year traveled. I happened to read a quote elsewhere showing that there is no aging in that case.

1 more question: Do things that move slower than the speed of light follow classical time calculations or is time relative for them, too? E.g., in our first thought experiment, if it is an electron that is shot instead of a photon, I realize that the speed of the electron will be slower and the actual equations must be modified. But is there some kind of switch for "light = SR and anything else = classical calculation"?
(This probably seems like a stupid question. The reason I'm asking it is because of a passage I read that said something like: "If someone throws a baseball at you at 100mph and you are standing, it is approaching you at 100mph. If you run away at 20mph, it approaches you are 80mph. The speed of light, however, approaches you at 670,616,629mph if you are stationary and, even if you are running away at 20mph, it still approaches you at 670,616,629mph. So basically my question is if something is moving at, say, 670,000,020mph and you run away from it at 20mph, at what speed is it approaching you?)

Last edited: Jul 5, 2015
13. Jul 5, 2015

### Staff: Mentor

Bob can't travel exactly at the speed of light. But he can travel very, very close to it (if we leave out all the practical issues you said you were leaving out). The closer to the speed of light he travels, the less he ages; by making his speed close enough to the speed of light, he could make the aging he experiences on the trip from Earth to Planet X as small as you like.

However, when Bob arrives on Planet X, he will find that Planet X has advanced in time at least 1,000 years during his trip. (This assumes that Earth and Planet X have synchronized their clocks and calendars, so Bob has some way of telling how much time has elapsed on Planet X.)

14. Jul 5, 2015

### Staff: Mentor

No. Everything uses the SR calculation. The classical calculation is a good approximation to the SR calculation for speeds much less than the speed of light, but obviously that doesn't apply in the scenarios you're proposing.

See the relativistic velocity addition law:

15. Jul 5, 2015

### Janus

Staff Emeritus
Here's the trick everything is subject to Relativity, even your 100 mph and 20 mph question. To be absolutely correct, you would not measure the speed of the ball as being 80 mph with respect to you, but 80.000000000000355771656721874968 mph or ~ 1/44362242 of an inch per hour faster. Now it is unlikely that you were even able to measure the original 100 and 20 mph that accurately, so in a practical sense this correction is pointless.

So while these relativistic effects do technically exist at every day speeds, they are just too tiny to pay any attention to. We can just assume that we can subtract 20 mph from 100 mph and get an answer, that while not perfectly accurate, is accurate enough for all practical purposes.

There is no defined "switch over point" where you would stop being able to use classical calculation and start using SR, as this would depend on the situation and the degree of accuracy needed.

16. Jul 5, 2015

### Stephanus

Dear Janus, JustinM523.
This is the source of mine, also. Janus, is this a language problem that I see?
Which one is the correct statement?
A. This also means that the time (Person 2, standing near the start clock) reads P1's clock at the moment the light left is 0.577 sec
B. This also means that the time P1 reads P2'clock at the moment the light left is 0.577 sec.
C. This also means that the time P1 sees P1's clock at the moment the light left is 0.577 sec.
But judging by logic, and by your ST diagram #11, statement A is true. Statement C is also false, I think, because at 0.577 sec, P1 does not know YET, that the light has left P2, but by 1.7 sec then P1 sees the light reaches P1, and at that time P1 sees P2 clock is 1 sec.
What Peter Donis says is true, "classical calculation is a good and 'practical' approximation" of SR. The formula of velocity addition is $w = \frac{u+v}{1+\frac{uv}{c^2}}$
Question A, baseball approaching you at 100mph while you run for 20mph, unfortunately the ball doesn't hit you at 80mph but...
Let speed of ball, u = 100mph, or 1.49116E-10 the speed of light.
Your speed, w = 20mph, or 2.98232E-11 the speed of light.
You see here, that u and w compared to the speed of light is neglible, so we won't have a significant difference from 'classical physics'.
Speed of ball hits you is v, so
$w = \frac{u+v}{1+\frac{uv}{c^2}}$
$v = \frac{w-u}{1-uw}$
$v = \frac{2.98232E-11 - 1.49116E-10}{1-4.44712E-21}$
$v = -80.000000000000mph$, hmmhhh, actually it should be smaller than -80mph, not -79.999... but at -80.0000 ..... 1. The calculator can't calculate that.
Or in absolute number, the ball actually hits you slightly more than 80mph.
the difference is negligible., so, you see in our everyday life, things just seems like in 'classical physics', but not GPS!
Okay, Question B: you run at 20 mph, and something approaching you at 670,000,020mph, so... It will hit you at 670,000,000.000258684, not 670,000,000
Using just the calculation above.
I think the most significant thing is the instant acceleration.
If we use gradual acceleration like in every day life to reach the speed of light, what we should use?
Is it GR problem or SR with differential integral?

17. Jul 5, 2015

### Stephanus

What?!! Janus has already answered that? And in more accurate floating points! Hmmhh...