# Spaceship at speed of light

1. Nov 2, 2015

### abcjake

Say my spacecraft is traveling c, and I'm walking .5 m/s towards the front of the spacecraft. Would that not mean I am traveling faster than light? I'm sure everyone has seen movie where a ship is traveling at the speed of light, while someone is moving about the cabin. I know only photons can only travel at light speed, but hypothetically speaking, wouldn't it be impossible to live while traveling c? You're heart can't pump itself any more forward, or it would be traveling faster than light.

I know it's a stupid question, but I can't seem to find the answer anywhere else online.

2. Nov 2, 2015

### BvU

Hi Jake, welcome to PF !

Not a stupid question. You don't want to know how often it pops up in one form or another.
The difficulty is in getting up to this magical speed you mention. Core of the special relativity theory is that nothing with mass can reach that speed. With good reasons.

So we make a little step back: suppose you manage to almost get up to speed. Say 0.99999999 c . And then you throw a ball or something. With a speed of 10 m/s relative to you. Who would there be to say it goes faster than c ? Some bloke left back on earth (if that's where you started from) ?

No, that guy would conclude a speed of 0.9999999901 c or so. There are (non-trivial) addition rules for speeds that feature this magical speed c as an upper limit.

Last edited: Nov 3, 2015
3. Nov 2, 2015

### DaveC426913

As BvU points out, you cannot travel at c, but you can get arbitrarily close to it.
And that's not splitting hairs. Here's why.

First, remember that your speed is always relative to something else. You, in your spaceship, always perceive yourself to be stationary. Your clocks all tick at one second per second. Your flashlights all send out rays of light moving at c, regardless of which way to point them. So, your speed is relative to some reference point - let's say it's Earth.

At BvUs' suggestion you are moving at 0.99999999c relative to Earth. Your time dilation factor is a little over 7000:1. That means, from Earth's point of view, you are almost frozen inside your spaceship, they see you moving at 1/7000th of normal. Your heart beats once every 12 hours. So, when you take a step, it is not .5m/s, it is actually, according to Earth, .5m every 7000 seconds, or about 4m per hour. This additional speed gets you nowhere near 1.0c, more like .999999990001c.

4. Nov 2, 2015

### Janus

Staff Emeritus
First you have to realize that velocity is not absolute, but relative. So when you say you are traveling at c (or better near c) you have to consider what you are moving relative to. So for instance, you could say that you are moving at near c relative to the Earth. But the flip side is that you can say that the Earth is moving at near c relative to you. What it comes down to is that there is no preferred reference by which to judge motion, and there is no test that you could perform inside your ship that can distinguish between your ship moving at near c or standing still.

I'll give you a quick example using just Newtonian physics. Say you are in a train sitting at a station and you begin to walk forward at 0.5m/s and you mass 70 kg. Your kinetic energy with respect to the ground will be 0.52x 70/2 = 8.75 joules, and that is the energy you will have to expend to go from standing still to moving at 0.5 m/s.

Now imagine that the train is moving at 100 m/s with respect to the ground. Standing still in the train, you now have a KE of 1002x 70/2 = 350000 joules with respect to the ground. If you now start to walk forward at 0.5 m/s with respect to the train, your new KE with respect to the ground will be 100.52x 70/2 = 353508.75 joules which means it increased by 3508.75 joules. This is some 401 times more than the 8.75 joules you expended when the train was standing still. However, you do not need to expend 401 times as much energy to walk at 0.5 m/s when the train is moving than you do when it is standing still. (think about it, does it take more effort to reach forward and change the setting on your stereo in the car while it is driving down the freeway than it does when sitting in the driveway/). If fact, as far as you are concerned, it still only takes an output of 8.75 joules. The amount of energy it takes to walk forward in the train does not change just because the train is moving.

Now that's dealing with Newtonian physics, With Relativity, there are a few more wrinkles. For one, You, in the moving train, do not measure time and space the same as the person on the ground does. As a result, if you start walking forward at 0.5m/s in the train as measure by you, the person on the ground will not measure your speed as being 100+0.5 = 100.5m/s, but a bit less than that (granted, at these speeds the difference will be so exceedingly tiny that anyone would be hard pressed to measure it. This is why at everyday speeds we can still get away with using the simple velocity addition method.). If you increase the speeds, the difference becomes larger and larger, and when you start getting into appreciable fractions of the speed of light, they start to become readily apparent. The other thing is, that, no matter what the speeds are, they will never add up to or exceed the speed of light. So for example, if the train is moving at 0.75c and you are walking at 0.75c relative to the train, then the person standing on the ground will only measure your speed as being 0.96 relative to himself. ( and you will measure your speed as being 0.96 relative to him.) Again, this is due to the difference between how he and you measure distance and time due to your relative speed.

So you can very easily walk forward at 0.5 m/s in a spaceship traveling at nearly c relative to some reference with no trouble on your part and not exceed c as measured with respect to that reference.

5. Nov 2, 2015

### abcjake

Ok, that makes more sense. Please excuse my stupidity, I'm still in HS taking physics and have a very limited knowledge of relativity. Let's say I have 3 objects in space forming a straight line. The center remains stationary while the other two on the left and right fly in opposite directions at near c, with respect to the center object. Considering that the two objects are now traveling opposite of each other at near c, with respect to the stationary object in the middle, is it still impossible to say that they are traveling faster than c relative to each other?

My guess would be, for example: the object on the left traveling near c would view the remaining objects behind it standing still because the light of the right hand object or center object cannot reach it, therefore not moving at or faster than c with respect to the right hand object. Is this a correct assumption?

Last edited: Nov 2, 2015
6. Nov 2, 2015

### Janus

Staff Emeritus
Okay, one of the basic precepts of Special Relativity is that you will always measure light moving at c relative to you regardless of the relative motion of the source with respect to you. That means that the light from any of these spaceships can be seen by any of the other ships. (imagine that at some point the leftmost ship is 1 light hr from the rightmost ship as measured by the rightmost ship. At that time, the leftmost ship emits a flash of light. From the rightmost ship point of view that light will travel towards him at c from that point and reach him in 1hr while the ship that emitted it will travel away from that point at near c. (note that this means that from the rightmost ship's point of view this means that the difference in speed between the light coming towards him and the ship moving away is going to be larger than c. This is okay in Relativity, because it only requires that nothing can move faster than c relative to you as measured by you.
On the other hand, from the leftmost ship's point of view, the same light travels away from him at c while the rightmost ship moves away from him at near c. The light has to chase after the rightmost ship in order to catch up with it, and will not reach it in 1hr. he also will measure the difference in speed between the light and the rightmost ship as being less than c. Again this is due to the two ships measuring time and space differently.

As far as what the ship's see of each others speed: Let's say that the rightmost and leftmost ships are traveling at 0.99c away from the center ship as measured by the center ship. Then the rightmost ship will measure the following: He will measure the center ship as moving away at 0.99c and the leftmost ship moving away at ~0.99995c or just 0.00995c faster than the leftmost ship.

The leftmost ship will measure the center ship as moving at 0.99c and the rightmost ship as moving at ~0.99995 c.

Again let's look at this in terms of everyone measuring the speed of light as c relative to themselves. We will start with all three ships in the same spot. At the the two ship leave at 0.99c a flash of light is emitted from that spot.
If we look at things from the center ship's point of view the following happens:
The light expands outward at c from him with the two ships chasing after it at 0.99c, since the light is moving faster the ships cannot catch up to the light front.
Now consider what happens the rightmost ship. He measures the speed of light as being c with respect to himself. Thus he sees the light expanding outward at c with himself at the center. The center ship is moving away at 0.99 c chasing after the light and never catching it. But what about the leftmost ship? We already established that it never catches the light according to the center ship. And we can't have it catching up to and passing the light according to one ship but not the other. (for example we can imagine a scenario where if the leftmost ship does catch and pass the light front, it sets off a bomb on the ship destroying it. We can't have the leftmost ship blowing up according to the rightmost ship but not doing so according to the leftmost ship.) So in order to keep things consistent between ships, the rightmost ship cannot measure the leftmost ship as moving faster than c relative to him. The same happens when we look at things from the leftmost ship's perspective. it sees itself at the center of the expanding light and the rightmost ship always moving at less than c in order to stay in that expanding sphere of light.

7. Nov 3, 2015

### BvU

No need to apologize. You ask good questions. Questions some folks don't even ask at university level -- and afterwards they can't effectively get their heads around relativity issues for a whole life !

8. Nov 3, 2015

### Mister T

You can start with the Principle of Relativity. You can easily find the excellent account written by Galileo of being in the hold of a ship. Once you learn that there is no difference between uniform motion and rest, then move on to the second postulate of Einstein's relativity. It's an assertion that tells us no matter how fast we travel chasing a light beam, the beam will recede from us at the same speed as it did when we were not chasing it. There's a quip about understanding this idea, and it goes like this. If you think you understand it when you first encounter it, then you don't.

Let's refer to them as A, B, and C. B is in the middle. The speed of A with respect to B is 0.8 c. Likewise, the speed of C with respect to B is 0.8 c.

Paradoxically, B will see the separation distance between A and C increase at a rate of 1.6 c.

But A will see C recede at a speed of about 0.98 c. Likewise, C will see A recede at at a speed of about 0.98 c.

The easiest way to do the math is to introduce the speed parameter $\theta$ defined by the equation

$$\beta=\tanh \theta.$$
In this example $\beta=0.8$. Use the above relation to find that $\theta \approx 1.0986$. Instead of adding 0.8 + 0.8 to find the relative speed of A with respect to C, we add 1.0986 + 1.0986 to get 2.1972. Then again use the above relation to find $\beta$.

$$\beta=\tanh (2.197) \approx 0.98.$$
So in Galilean relativity we add the speeds $\beta$ but in Einsteinian relativity we add the speed parameters $\theta$.

This is an example of the use of hyperbolic spacetime geometry.

Last edited: Nov 3, 2015
9. Nov 5, 2015

### abcjake

Thank you for the detailed response! Your explanation makes a lot more sense now.