Effects of time dilation for near-speed-of-light travel

[Chenkel]

TL;DR Summary

In this post I attempt to present my interpretation of the effects of time dilation for space travel, I'm looking for feedback and corrections.

Hello everyone,

I've been learning about special relativity, and so far I believe based on what I read that if you are traveling at a velocity of .6c, you will experience time 20 percent slower than people on earth.

Each second in the spaceship will be 1.25 earth seconds.

Each second on earth will be .8 spaceship seconds.

Hopefully I'm correct in my analysis so far.

I'm wondering, if a galaxy is determined to be 10 light years away from a person on earth, does that mean that the person in the spaceship will get there in 8 spaceship years? Furthermore does it also means that the person on earth will know based on the "math" that the traveler got there after 10 earth years have passed since launch of the spaceship?

If the answers to these two previous questions are "yes" am I correct in saying that the person on earth will actually be able to receive confirmation with a powerful enough telescope after 18 light years have passed that the spaceship person arrived at the galaxy since the launch of the spaceship?

My reasoning goes like this, if the spaceship has gotten to the destination, it traveled 10 light years according to earth calculations, but the person in the spaceship views the journey as 8 light years when traveling at .6c, therefore because it was only able to send a signal to earth after 8 spaceship years, and the journey relative to a receiver of the signal on earth takes 10 years, then the first signal will be received in 18 years.

So if one earth second passes when .8 spaceship seconds pass, and light is moving at the same distance in meters per second regardless of reference frame, does that mean that the length of space in front of the spaceship is "shrunken" down relative to earth meters?

Finally my last question which I'm very interested in is the following, if the spaceship is able to do these trips at near speed of light, doesn't it mean that you can majorly reduce the amount of time that a person is subjected to during a space journey through time dilation alone?

I'm wondering how to calculate the effective
reduction in time based on time dilation alone and to see how numerically significant it is, but I think I need to understand the math a little better to do that.

I appreciate if anyone has any insights to offer, and can perhaps show me if I am doing something wrong in my analysis, so far the math seems to make sense, but I'm not sure if I'm missing something.

Let me know what you think, thank you!

 

[Nugatory]

am I correct in saying that the person on earth will actually be able to receive confirmation with a powerful enough telescope after 18 light years have passed that the spaceship person arrived at the galaxy since the launch of the spaceship?

No. Travel time is distance divided by speed so 10/.6 is not quite 14 years for the spaceship to make it to the distant star, plus ten years for the light signal to get back to us. What the traveler’s clock reads is irrelevant to the earth person’s clock readings.

 

[Chenkel]

No. Travel time is distance divided by speed so 10/.6 is not quite 14 years for the spaceship to make it to the distant star, plus ten years for the light signal to get back to us. What the traveler’s clock reads is irrelevant to the earth person’s clock readings.

My bad, thank you for pointing that out, I'm not sure how I missed that.

 

[Chenkel]

No. Travel time is distance divided by speed so 10/.6 is not quite 14 years for the spaceship to make it to the distant star, plus ten years for the light signal to get back to us. What the traveler’s clock reads is irrelevant to the earth person’s clock readings.

I reworked the math.

10 light years away is a measure of distance to the galaxy.

The spacecraft travels at v = .6c

So in roughly 10/.6, or 16.66 earth years the spaceship will get there.

But because time dilation, the trip will feel like .8*16.66 in spaceship years, or 13.328 spaceship years.

Am I correct in my analysis?

The effect from time dilation seems possibly very significant in and of itself, it would be very cool if there was a way to calculate how many years could be shaved off a journey as a function of the time dilation alone.

 

[FactChecker]

if you are traveling at a velocity of .6c, you will experience time 20 percent slower than people on earth.
Each second in the spaceship will be 1.25 earth seconds.
Each second on earth will be .8 spaceship seconds.
Hopefully I'm correct in my analysis so far.

One must be very careful to keep track of which observer he is talking about for any observation.
This is correct from the Earth observer's perspective, but not from "yours". Instead of "you are traveling" with respect to the Earth, you can legitimately think of yourself as the stationary one and the Earth is traveling with respect to you. From your perspective, it is the Earth's clock and processes that are slowed down. Each of your seconds is 0.8 Earth seconds and each Earth second is 1.25 of your seconds.
(Remember that "you" and an Earth observer are going in opposite directions with respect to each other. As time passes, "you" and the Earth observer have traveled a long distance in opposite directions and are observing clocks and processes in opposite directions. How each has synchronized his clocks over those distances makes all the difference.)

 

[Chenkel]

One must be very careful to keep track of which observer he is talking about for any observation.
This is correct from the Earth observer's perspective, but not from "yours". Instead of "you are traveling" with respect to the Earth, you can legitimately think of yourself as the stationary one and the Earth is traveling with respect to you. From your perspective, it is the Earth's clock and processes that are slowed down. Each of your seconds is 0.8 Earth seconds and each Earth second is 1.25 of your seconds.
(Remember that "you" and an Earth observer are going in opposite directions with respect to each other. As time passes, "you" and the Earth observer have traveled a long distance in opposite directions and are observing clocks and processes in opposite directions. How each has synchronized his clocks over those distances makes all the difference.)

How do you determine for which observer the time has actually slowed?

If a body's kinetic energy is indicative of near light speed travel, and another body has less kinetic energy, can you assume the body with less kinetic energy is the observer that is moving through time at quicker pace?

 

[phinds]

if you are traveling at a velocity of .6c, you will experience time 20 percent slower than people on earth.

Just to be sure you understand, this is not correct. You will experience time at the rate of one second per second (your proper time) and a person on Earth will experience time at one second per second (his proper time). You will each see the other's time slowed down and neither of you will experience any slowing down of your time.

 

[FactChecker]

How do you determine for which observer the time has actually slowed?

Each observes the other's clocks and processes to have slowed. There is no contradiction in that because they are going in opposite directions from each other and are observing clocks and processes in totally opposite directions.

 

[phinds]

Oh, and just to somewhat pile on to what has already been said, you really need to keep in mind that there are 4 things that need to be kept separate based on the two frames of reference, yours and Earth's

What you experience
What Earth sees you experiencing
What Earth experiences
What you see Earth experiencing

 

[Chenkel]

Oh, and just to somewhat pile on to what has already been said, you really need to keep in mind that there are 4 things that need to be kept separate based on the two frames of reference, yours and Earth's

What you experience
What Earth sees you experiencing
What Earth experiences
What you see Earth experiencing

Isn't the what the spaceship experiences and what the earth experiences different?

Let me give the following example, let's say two people live for 100 years exactly, and one person is on earth, and the other goes to a distant galaxy at v = .6c, wouldn't biological processes dictate that after the time the person on earth dies, the person in the spaceship will live an additional 20 years in the journey?

I am not sure I accounted for how the spaceship views earth, or how the earth views the spaceship, but isn't the way the spaceship views the spaceship, different than the way the earth views the earth?

 

[FactChecker]

Isn't the what the spaceship experiences and what the earth experiences different?

Each observer thinks that his clocks and processes are completely normal, as though he was stationary.

Let me give the following example, let's say two people live for 100 years exactly, and one person is on earth, and the other goes to a distant galaxy at v = .6c, wouldn't biological processes dictate that after the time

Whose "time" are we talking about here?

the person on earth dies, the person in the spaceship will live an additional 20 years in the journey?

20 years according to whose clocks?

I am not sure I accounted for how the spaceship views earth, or how the earth views the spaceship, but isn't the way the spaceship views the spaceship, different than the way the earth views the earth?

No. Either one is completely valid if he considers himself to be the stationary one. The principle of relativity says that all physical processes in an inertial reference frame behave as though that inertial reference frame was stationary.

 

[Chenkel]

Just to be sure you understand, this is not correct. You will experience time at the rate of one second per second (your proper time) and a person on Earth will experience time at one second per second (his proper time). You will each see the other's time slowed down and neither of you will experience any slowing down of your time.

I think I might understand what you are saying, but the biological processes must slow down for the person in the spaceship compared to people on earth.

One second on earth will be felt in the time that one second is felt on the space ship, but the "one second" on earth is defined as 1.25 spaceship seconds, and the "one second" on the spaceship is interpreted as .8 earth seconds.

The way I look at it is this, I imagine two computers that process information at 1000 bits per second, during one second experienced by the computer on the spaceship, the computer on earth will do 1250 operations.

During one second experienced by the computer on earth the spaceship computer will perform 800 operations.

I'm trying to intuitively understand what you are saying, proper time is a new concept to me and I'm pretty sure I'm not understanding something here.

 

[phinds]

I think I might understand what you are saying, but the biological processes must slow down for the person in the spaceship compared to people on earth.

This is exactly what I am saying is NOT correct.

Let's look at it this way. You, right now, are time dilated by a large factor relative to a particle in the CERN accelerator, a small amount relative to a passing asteroid, and a zero amount relative to the chair you are sitting in. Each of these sees your biological process proceeding at different rates. How could they possibly all be right? Also, do you feel any different because they see you as being different?

 

[FactChecker]

I think I might understand what you are saying, but the biological processes must slow down for the person in the spaceship compared to people on earth.

There is never a direct side-by-side comparison unless the traveling person returns to the original spot. That would require the "traveling" observer to turn around and not remain in his original inertial reference frame.

Going back to the case where both observers remain in their original inertial reference frame, suppose the two observers are side-by-side at the beginning of a second. One second later (in each observer's clocks), they are widely separated in opposite directions. So any comparison depends on how each has synchronized clocks (or physical processes) in opposite directions.

 

[Chenkel]

There is never a direct comparison unless the traveling person returns to the original spot. That would require the "traveling" observer to turn around and not remain in his original inertial reference frame.

Going back to the case where both observers remain in their original reference frame, suppose the two observers are side-by-side at the beginning of a second. One second later (in each observer's clocks), they are widely separated in opposite directions. So any comparison depends on how each has synchronized clocks (or physical processes) in opposite directions.

There is never a direct comparison unless the traveling person returns to the original spot. That would require the "traveling" observer to turn around and not remain in his original inertial reference frame.

If 10 different spaceships launch from earth at varying near-light speeds, and come back, I thought relativity says each person on each spaceship will be younger than earth compared to when they first launched because each spaceship clock that is compared with a singular earth clock will have ticked at a slower rate than the earth clock during the journey depending on how fast that particular space craft was going.

Why can't we use this to assume that one way trips slow down physical processes when compared to the way they would work on earth when not traveling near speed of light?

I do feel I'm missing something, but I'm not sure what it is, my logic seems to make sense to me but that doesn't mean I'm correct and I'm open to learn where I am not understanding correctly.

 

[FactChecker]

If 10 different spaceships launch from earth at varying near-light speeds, and come back, I thought relativity says each person on each spaceship will be younger than earth compared to when they first launched because each spaceship clock that is compared with a singular earth clock will have ticked at a slower rate than the earth clock during the journey depending on how fast that particular space craft was going.

Yes. If they return to the Earth, you are correct. In that case, the comparison at the end is side-by-side. The travelers have not remained in their original inertial reference frame and the problem is more complicated from their perspective. It is still a simple problem from the Earth observer's perspective.

Why can't we use this to assume that one way trips slow down physical processes when compared to the way they would work on earth when not traveling near speed of light?

In that case, there is no side-by-side comparison at the end. Without a return, you are starting at one location and ending at a location far away. So there is an issue of how clock times (or physical processes) are synchronized at a distance. The two inertial reference frames ("stationary" and "moving" assigned either way) disagree on how clocks are synchronized at a distance. That is called the "relativity of simultaneity".

I do feel I'm missing something, but I'm not sure what it is, my logic seems to make sense to me but that doesn't mean I'm correct and I'm open to learn where I am not understanding correctly.

Consider how light can appear to travel at the same speed in two different inertial reference frames that are moving with respect to each other. Start with a flash of light at time=0 for both clocks in one location and see how their clocks at a distance must disagree. Experimental results forced Einstein to consider that. Each observer thinks that his own clocks are normal and perfectly synchronized at any distance. Also notice that, since the two reference frames are going in opposite directions with respect to each other, they will each observe the other's clocks to be wrong (in the same way).

 

[phinds]

If 10 different spaceships launch from earth at varying near-light speeds, and come back, I thought relativity says each person on each spaceship will be younger than earth compared to when they first launched because each spaceship clock that is compared with a singular earth clock will have ticked at a slower rate than the earth clock during the journey depending on how fast that particular space craft was going.

NO NO NO !!!
clocks tick at one second per second. The reason for the difference in aging is NOT because their clocks tick at different rates (they do NOT) it's because they have taken different paths through spacetime

It's just like 2 cars taking a trip from Boston to DC and each going exactly 60MPH for the whole trip but taking path and therefore having different amounts of distance and time because of the differing paths.

 

[PeroK]

I do feel I'm missing something, but I'm not sure what it is, my logic seems to make sense to me but that doesn't mean I'm correct and I'm open to learn where I am not understanding correctly.

The thing you are missing is that differential ageing is different from time dilation. Velocity-based time dilation is a coordinate effect that is completely symmetric. It makes no sense to try to say whose time is "really" slower.

Differential ageing is asymmetric physical effect due to two objects taking different paths through spacetime. The proper time each experiences is equal to the length of their path through spacetime.

Unfortunately, you already seem to have "learned" that the traveller's time is really going slower than Earth time. You need to unlearn this!

 

[Ibix]

10 light years away is a measure of distance to the galaxy.

A minor point, but 10ly is barely past the nearest star. It's more like a million to the next galaxy.

The key to your problem is to realise that there are very important differences between a journey where one travels from A to B and one where one also returns to A. The main one here is that if you return then you can directly compare your clocks to clocks that remained at A, and there has to be a unique answer to which one shows more elapsed time. If you do not return, however, you can only look at the other person's clock through a telescope, and you will see it as it was several years ago. You can, of course, calculate what it would be showing today by adding the distance divided by the speed of light - but different frames measure different distances. So they have different ideas of "what time it is over there right now". And if you work through that in detail it means that both the person travelling from A to B and the person staying at A agree that the other is younger. (That simple analysis fails when someone turns round unless you are careful, which is why this doesn't contradict the "single result when one returns home".)

 

[Chenkel]

NO NO NO !!!
clocks tick at one second per second. The reason for the difference in aging is NOT because their clocks tick at different rates (they do NOT) it's because they have taken different paths through spacetime

It's just like 2 cars taking a trip from Boston to DC and each going exactly 60MPH for the whole trip but taking path and therefore having different amounts of distance and time because of the differing paths.

I think I might see what you are saying, I feel I have a little intuition with the second paragraph, but the first paragraph confuses me a little, if a rocket leaves earth and comes back it's not clear to me how the homebody and the traveler are not interchangeable, it seems as long as you ignore acceleration, all movement in any direction can be seen as relative, and so the stationary person can be interpreted as a traveler or vice versa.

On the subject of proper time. If someone travels in (10 seconds * v) meters, where v is equal to .6c, then the traveler counts his "own" seconds at the same rate the homebody counts the home body's "own" seconds; hopefully I'm correct in saying that when a reference frame measures its own clock it's reading proper time.

However if I'm not mistaken the reference frame that is treated as moving is imagined to be ticking slower by the reference frame chosen to be stationary.

The traveler is interchangable because the derivation of the Lorentz equation for time dilation is a geometric symmetry where either observer is interchangable as the traveler.

This means that both reference frames see the clock in the "other reference frame" as ticking slower.

To make a basic example suppose that OS represents the reference frame of the observer who is stationary, and OT represents the reference frame of the observer who is traveling.

If the traveler travels for 10 seconds proper time at v = .6c then he will travel a distance of d = (10 seconds) * .6c.

After 10 seconds the traveler will have reached the "destination" and he will have to return home, but at 10 seconds in proper time according to OS's reference frame OT's clock will be estimated to be at 8 seconds (in OS's reference frame) due to time dilation, but the journey back will be 10 seconds proper time.

Because of this when OT returns to OS, OS's clock will tick 20 seconds in proper time during the round trip journey, and in OS's frame of reference, OT will be perceived to tick 16 seconds.

Now this at first seems to makes sense to me, but what about this..

If we swap OT and OS by changing the stationary reference frame, which we should be able to do by symmetry, then OT will now perceive OS's clock to have ticked 16 seconds, and OT will now see OT's clock as having ticked 20 seconds.

How can this seeming inconsistency be explained?

 

[PeterDonis]

@Chenkel, please read this:

https://www.physicsforums.com/threads/when-discussing-the-twin-paradox-read-this-first.1048697/

 

[PeroK]

How can this seeming inconsistency be explained?

Suppose A travels for a time $t_a$ according to her clock and observes B turn round when B's clock reads $t_b$.

What does B's clock read when B observes A turn round and what does A's clock read at this point?

Can you use this to determine who "really" turned round? Assuming the usual twin scenario.

 

[malawi_glenn]

This means that both reference frames see the clock in the "other reference frame" as ticking slower.

you also need to account for the fact that light has to travel from the clock at the spaceship to the person on earth, and vice versa. In order to calculcate the total amount of "time lagging".

 

[Chenkel]

@Chenkel, please read this:

https://www.physicsforums.com/threads/when-discussing-the-twin-paradox-read-this-first.1048697/

Thank you for the article! I will study it.

 

[Chenkel]

Suppose A travels for a time $t_a$ according to her clock and observes B turn round when B's clock reads $t_b$.

What does B's clock read when B observes A turn round and what does A's clock read at this point?

Can you use this to determine who "really" turned round? Assuming the usual twin scenario.

Does propertime(t_a) always equal propertime(t_b)?

Also if A sees B turn around, doesn't that mean A already turned around?

I'm not sure what you mean by "turn around."

What are you trying to show by your example?

 

[FactChecker]

I think I might see what you are saying, I feel I have a little intuition with the second paragraph, but the first paragraph confuses me a little,

In SR, there is a trade-off between travel through 3-dimensional distance and travel through time. A moving observer travels through 3D distance and so travels less through time. His clock ticks at exactly the right rate for that travel through time, but there is less travel through time.

if a rocket leaves earth and comes back it's not clear to me how the homebody and the traveler are not interchangeable, it seems as long as you ignore acceleration, all movement in any direction can be seen as relative, and so the stationary person can be interpreted as a traveler or vice versa.

As long as there really is no acceleration, you are right. But if there is acceleration, you can not ignore it. All inertial reference frames have the same physics, so unaccelerated motion can be considered relative and there is symmetry. But that is not true if there is acceleration. You must not confuse the mathematical second derivative of position with physical acceleration. The first is symmetric, but the second is not.
In physics, it is not true that ALL motion is relative, only unaccelerated motion.

 

[PeroK]

Does propertime(t_a) always equal propertime(t_b)?

No!

Also if A sees B turn around, doesn't that mean A already turned around?

How do you figure that out?

I'm not sure what you mean by "turn around."

Change direction.

What are you trying to show by your example?

The asymmetry inherent in the scenario that your casual analysis has overlooked!

 

[Chenkel]

In SR, there is a trade-off between travel through 3-dimensional distance and travel through time. A moving observer travels through 3D distance and so travels less through time. His clock ticks at exactly the right rate for that travel through time, but there is less travel through time.

As long as there really is no acceleration, you are right. But if there is acceleration, you can not ignore it. All inertial reference frames have the same physics, so unaccelerated motion can be considered relative and there is symmetry. But that is not true if there is acceleration. You must not confuse the mathematical second derivative of position with physical acceleration. The first is symmetric, but the second is not.
In physics, it is not true that ALL motion is relative, only unaccelerated motion.

What is the difference between the second derivative of position, and physical acceleration? I thought these were the same.

 

[Chenkel]

No!

How do you figure that out?

Change direction.

The asymmetry inherent in the scenario that your casual analysis has overlooked!

Are you saying that because my example doesn't take into account acceleration (the process by which one turns around) it therefore leads to an asymmetry that seems contradictory?

 

[FactChecker]

What is the difference between the second derivative of position, and physical acceleration? I thought these were the same.

When the coordinate system is rotating or accelerating, the second derivative in that coordinate system is very different from physical acceleration. Here is an extreme example.
Suppose a coordinate system is attached to the head of a pilot in a jet airplane with the center in his head and the x-axis pointing forward out his nose. Suppose he is looking forward and the plane is flying at Mach one. If he turns his head to look sideways, the derivative of x goes from Mach 1 to 0 in about one second. That would be a hell of a lot of acceleration! So there can be a huge difference between the second derivative and true physical acceleration.

 

[Ibix]

What is the difference between the second derivative of position, and physical acceleration? I thought these were the same.

Not in general. They are numerically equal if you use an inertial frame, but consider a non-inertial frame that you might use in an accelerating lift. You are in a constant position, yet any accelerometer will show that you are accelerating.

To be clear, the kind of acceleration you measure with an accelerometer is called "proper acceleration". The second derivative of your position coordinate (in whatever frame you choose to use) is called "coordinate acceleration".

 

[Ibix]

Are you saying that because my example doesn't take into account acceleration (the process by which one turns around) it therefore leads to an asymmetry that seems contradictory?

No - quite the opposite. The two people have different experiences, so "everything's the same for both of us, so we both exect the other to be younger" is not valid reasoning.

In fact, the problem is that the naive analysis tries to use the traveller's outbound inertial rest frame for half the problem and the inbound frame for the other half. That doesn't work without some care and extra maths that the naive analysis neglects.

 

[Chenkel]

No - quite the opposite. The two people have different experiences, so "everything's the same for both of us, so we both exect the other to be younger" is not valid reasoning.

In fact, the problem is that the naive analysis tries to use the traveller's outbound inertial rest frame for half the problem and the inbound frame for the other half. That doesn't work without some care and extra maths that the naive analysis neglects.

Perhaps I'm mistaken, but it does sound like you are maybe agreeing with me on why the analysis I wrote doesn't work based on your insight.

My analysis uses a non accelerated reference frame during the first half, and then an accelerated reference frame during the second half, and I agree with you that it is wrong and naive without considering the relativistic effects of acceleration (it seems acceleration cannot be ignored based on what I've read, I'm not exactly sure why though), thus making a theoretical contradiction between what what the two observers see when they arrive at the same spot after the round trip journey.

I did say that there was some asymmetric (it's possible that was a bad choice of word on my part) relationship between what the two observers saw, statement 1 = clock A is younger and clock B is older, or statement 2 = clock A is older and clock B is younger; if I'm not mistaken, statement 1 and statement 2 cannot both be true simultaneously for proper time measurements, but one of them can be true or neither can be true for proper time measurements.

 

[PeterDonis]

Thank you for the article! I will study it.

Please do. And I would strongly suggest not posing any further questions in this thread until you have. Many of the questions you are asking will be answered if you read the article.

 

[PeterDonis]

What is the difference between the second derivative of position, and physical acceleration? I thought these were the same.

The second time derivative of position is frame dependent.

"Physical acceleration", which is usually called proper acceleration in relativity, is what an accelerometer attached to the object reads. It's an invariant, the same in all frames.

Generally, proper acceleration is what you want to focus on.