# Spaceships with relativistic instruments

I wonder if it is possible, for future spaceships, to have somekind of instruments on board, that will tell the pilot what his true speed through spacetime is and how much time dilation he is currently experiencing.

Let me explain:

Imagine an instrument, that at any time it can tell the pilot 2 numbers. These 2 numbers are in the form of percentages, and the first is about the pilot's speed in space, and the second is the pilots speed in time.

For example, if the instrument reads, [60% 80%] then your speed in space is 60% the speed of light, and your speed in time is 80% the speed of light (so you are experiencing a 100% - 80% = 20% time dilation).

If the instrument reads [100% 0%] then your speed in space is 100% C and your speed in time is 0% C - so basically your are traveling at the speed of light (something that of course can never happen). If the instrument reads [0% 100%] then your speed in space is 0% C and your speed in time is 100% - so basically you are in freefall inside a gravitational field.

If the clock reads for example [0.00001% 99.9999999999995%] then you are on Earth, moving at 0.00001% the speed of light (or at 29.9 m/s or ~ 100 km/h) in space and 99.9999999999995% C in time, etc...

Could such an instrument exist? Or it would be impossible due to relativity to have such an instrument? Do such instruments exist already? And if yes, how are they called?

• "true speed through spacetime" - are you supposing some absolute frame of reference? Commented Sep 8, 2022 at 14:00
• This could conceivably be done by measuring the red or blue shift of the Cosmic Microwave Background. The CMB functions as a de facto absolute global reference frame, even though locally there is no absolute frame. Commented Sep 8, 2022 at 14:32
• What do the phrases "true speed through spacetime" and "how much time dilation he is currently experiencing" mean? In mainstream physics they don't mean anything. If you are rejecting mainstream physics, your question is off topic. If you are inventing your own new private vocabulary for mainstream physics phenomena, then nobody can understand your question unless you provide a glossary. Commented Sep 9, 2022 at 0:49
• @nuke : if you weren't going to answer my questions, why did you type so.many words? Commented Sep 9, 2022 at 12:08
• Nuke: There you go again. You write "So It seems to me that I should be able to always tell what my time dilation is", but you refuse to tell us what "my time dilation" means. Either it means nothing at all (in which case you are talking nonsense) or you are refusing to tell us what it means (in which case you are intentionally wasting our time). Commented Sep 9, 2022 at 20:04

The problem is that in the spaceship's rest frame the spaceship is at rest. That means the spatial velocity is zero metres per second and the velocity in time is one second per second. So your instruments would always read 0% and 100%.

I'm guessing that what you mean is how your spatial and temporal velocities would be seen by an observer on Earth, i.e. your time meter should read 50% if an observer on Earth sees your time as dilated by a factor of $$\tfrac12$$. The problem is that this depends on the velocity of the Earth relative to you, or in general relativity also the gravitational potential of the Earth relative to you.

If you know these quantities you could calculate your time dilation as observed from Earth, but there is no way for your meters to show this without this external information. For example, suppose your spaceship had no windows so you couldn't see out and therefore couldn't see where the Earth is and how fast it is moving. Then there is no way your meters could show the information you want.

• Yes but relative to an inertial frame like the planet Earth, my spatial and temporal velocities should be somewhere between 0% and 100% but never 0% or 100%. Right? When I travel I want to know my speed through spacetime relative to my origin, which was the Earth. So maybe relative to that we can has such instruments?
– Nuke
Commented Sep 26, 2022 at 18:25

No, it is not possible. You do not 'experience time dilation' in the sense you suppose. Time dilation is relative, not absolute. In my frame of reference, a muon passing at close to the speed of light might be time dilated by a factor of 60, so that a second passes for the muon while a minute has passed in my frame. However, from the perspective of the muon it is I who is time dilated, and if my heart is beating once per second in my frame it is beating at once per minute in the frame of the muon. Both the muon and I experience time at the same rate, namely one second per second, in our own frames.

Your underlying misconception is that you think it is possible in some absolute sense to talk about your movement through space and through time. That assumes there is an absolute frame of reference, which as far as we know does not exist.

In the example I raised earlier, in the muon's rest frame, the muon is moving purely through time, and it is I moving through space. In my frame, I am moving only through time and the muon is moving through space. Essentially, I and the muon have our own individual time axes; I am always moving directly up my time axis, and the muon is always moving directly up its own time axis. But if the muon and I are moving relative to each other, our individual time axis are not aligned, but are tilted relative to each other.

Differences in overall elapsed times, of the sort that occur in the twin paradox, can be measured, but again they are relative, and importantly they arise as a result of changing the direction of one's time axis during periods of acceleration. When the travelling twin is coasting, on either the outbound or inbound leg of the journey, the stay-at-home twin is time dilated in the traveller's frames. It is only at the turn-around point that a difference in elapsed time accrues to make the travelling twin younger upon return. That effect is a result of the relativity of simultaneity, and is dependent on distance- it is not time-dilation in the usual sense, which is always symmetric.

• @PM2ring many thanks for the edit! Commented Sep 9, 2022 at 10:22

If "speed in space" is $$\frac{dx^i}{d\tau}$$ and "speed in time" is $$\frac{dt}{d\tau}$$ then this should be possible.

In Special Relativity,

$$d\tau = (dt^2 - \frac{ds^2}{c^2})^{\frac{1}{2}} = (1-\frac{v^2}{c^2})dt$$

$$\tau_2 - \tau_1 = \int_{\tau_1}^{\tau_2} (1-\frac{v^2}{c^2})^{\frac{1}{2}}dt$$

The easiest way I can think of to match this theory to something in reality is to imagine a rocket, with its own clock, in a three dimensional coordinate system where all the coordinates are synchronized to a single clock. The coordinate system has sensors that measure the movement of the rocket and transmit that information to the rocket and/or the rocket transmits its wristwatch time $$\tau$$ to the coordinate system observer. Actually, the rocket wouldn't have to measure $$\tau$$. The coordinate observer can calculate it using the formulas above (or something more complicated if there is a gravitational field or acceleration). So really, all you need is a synchronized grid that tracks the movement of an object.

You are thinking in the right direction, but you seem to have a few misconceptions. In special relativity, any observer will measure the speed through spacetime of anything as c, or $$1$$ in natural units. This is the magnitude of the velocity $$4$$-vector. It is invariant.

You are suggesting measuring the components of this $$4$$-vector and calculating the magnitude. This can be done with existing technology. You pick two points in spacetime and measure the spatial and time components of the separation between them. In principal you use a ruler and a clock. In practice for spacecraft, you use radar beams instead of a ruler.

Any observer will measure his own speed as purely temporal. I am sitting right here not moving. At $$t_0$$, my position is $$x_0$$. Later, at $$t_1$$, my position is still $$x_0$$. My $$4$$-vector displacement is $$(t_1−t_0,0,0,0)$$. The interval is

$$s_2=c^2(t_1−t_0)^2−(0^2+0^2+0^2)=c^2Δt^2$$

Since I am not moving in my frame, the proper time interval $$Δτ=Δt$$.

The $$4$$-velocity is $$v^2=s^2/τ^2=c^2$$

The question is what the pilot will measure for the $$4$$-velocity of another spacecraft or the earth. For those, he will likely discover both temporal and spatial components. The magnitude will still be $$c$$.