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Thomas Smith
Thomas Smith

Through Spacetime

I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension of space-time at 1 second per second as time goes forward.

Through Spacetime

Now, the statement about traveling through time 'at the speed of light' needs to be qualified. You can easily see that it does not make sense if you use ordinary definitions: the speed of light is measured in 'length per time', while a 'speed through time' would be measured by 'time per time', which is just a number.

However, we can make sense of this statement. We think of an observer as tracing a path through spacetime. To denote a point on this path we use a single coordinate that we call $\tau$. The path is defined by the functions $t(\tau)$ and $x(\tau)$: for each value of $\tau$ the observer is at a specific place $x$ at a specific time $t$.

Let's say we measure $\tau$ in seconds. We can now define a 'velocity vector' $u$ through spacetime, which is the rate at which $t$ and $x$ change when we change $\tau$: $$u=\left(c \fracdtd\tau,\fracdxd\tau\right).$$ Notice that I snuck a $c$ in there, to make sure $u$ has units of velocity.

The first component of $u$ is a nice definition of our velocity through time. The second component is some way of measuring velocity through space, but it is not the same as the velocity we usually think about, which is $\fracdxdt$.

Now comes a very nice mathematical theorem, which says that we can always assign values of $\tau$ to points on the path such that $u(\tau)=c$ at each point. With this choice of $\tau$, our 'velocity' $u$ is a constant, equal to the speed of light: no matter if we still or move very fast, our velocity through spacetime is the same. If we are sitting still, it just means that we are 'moving faster' in the time direction. If we are moving very fast (by the usual definition of moving...), then our velocity in the time direction will be smaller to compensate. Our only choice is where to point our velocity: a bit more in the time direction, or a bit more in the space direction.

It isn't. This idea seems to be something that the popularizer Brian Greene has perpetrated on the world. Objects don't move through spacetime. Objects move through space. If you depict an object in spacetime, you have a world-line. The world-line doesn't move through spacetime, it simply extends across spacetime.

Greene's portrayal of this seems to come from his feeling that because the magnitude of a massive particle's velocity four-vector is traditionally normalized to have magnitude $c$, it makes sense to describe the particle, to a nonmathematical audience, as "moving through spacetime" at $c$. This is simply inaccurate. A good way to see that it's inaccurate is to note that a ray of light doesn't even have a four-vector that can be normalized in this way. Any tangent vector to the world-line of a ray of light has a magnitude of zero, so you can't scale it up or down to make it have a magnitude of $c$. For consistency, Greene would presumably have to say that a ray of light "moves through spacetime" at a speed of zero, which is obviously pretty silly.

The reason we normalize velocity four-vectors for massive particles is that the length of a tangent vector has no compelling physical interpretation. Any two tangent vectors that are parallel represent a particle moving through space with the same velocity. Since the length doesn't matter, we might as well arbitrarily set it to some value. We might was well set it to 1, which is of course the value of $c$ in relativistic units. But this normalization is optional in all cases, and impossible for massless particles.

An object's total velocity through spacetime, its four-velocity, has a component in each dimension of spacetime (4). So, the four coordinates an object has in spacetime are $$x= \left( \beginarraycccct \\x^1(t) \\x^2(t) \\x^3(t)\endarray \right) $$ For the three dimensions of space, we see that these are just the positions as functions of the time as measured by whatever observer whose reference frame we are operating in. At the top, we see ct rather than t. Why? Well, t is measured in units of time, but we need it in units of length to make any sense out of the concept of spacetime. Alternatively, we can convert our positions into units of time by dividing by c. This is done often in astronomy, because the distances are so large (e.g. light-years, light-seconds, etc.).

We see that for a moving observer, $\gamma$ is greater than one, resulting in a larger velocity through time. This may seem counter-intuitive, since moving observers elapse less time. However, an intuitive way to think about it is this - velocity through space is how muc distance you can cover in some amount of time. So, velocity through time is the amount of time you can elapse in someone's reference frame in some amount of proper time. An observer moving at an enormous speed will record very little proper time, but an observer who he is moving with respect to will observe that it takes an enormous time for him to finish his journey. So, in a small amount of proper time our observer 'travelled' a very large distance through the coordinate time of the observer taking the measurements, hence a higher velocity through time.

Yes, the time-like component of the four-velocity of a stationary object is $c$. It's also correct that "everything moves at the speed of light through spacetime" -- this just means the magnitude of the four-velocity is $c$ (or rather, 1), and its direction keeps changing (note that the transformation here isn't really a rotation, it's a skew/boost, because of how the Minkowski dot product is calculated) as the vector slides on an invariant hyperbola (much like rotations slide on an invariant circle). This is not a matter of convention -- yes, you can choose other parameterisations for the worldline, but they don't satisfy the nice property of becoming equal to co-ordinate time when $v=0$.

Note that the other claim -- that the rest energy is a kinetic energy through time -- is wrong. This is trivially obvious, considering that everything moves at $c$ through spacetime, and the kinetic energy associated with a spatial speed of $c$ is infinity.

If you define "speed through space-time" as the Pythagorean sum of the speed through time (rate of proper time to coordinate time) and the speed through space, i.e. $\sqrt(\dfracc\Delta\tau\Delta t)^2+(\dfrac\Delta x\Delta t)^2$, then yes, every object in the universe has a spacetime speed of c (the speed of light) since this sum equals c for all objects.

Now, many commenters here seem to object to this definition of spacetime speed. But definitions are not true or false. Definitions are useful or not useful. You can say you don't like this definition of spacetime speed, but you can't say that it is wrong.

an invariant. So the absolute value of $\eta^\mu$ is always equal to c. If the mass has zero (spatial) velocity (the mass is standing still) the speed through time is $c$, and photons traveling through space only with speed $c$. In between these two extremes, the four-velocity consists of a part moving through time and a part moving through space.

"Frozen" word-lines, Block time, was a concept of Einstein and others ("the distinction between the past, present and future is a stubbornly persistent illusion" complemented by Weyl's commentary which one can read on page 101 in 2.7 on the Block universe section in the link to Weyl's commentary I made) implying that an object can't travel through space either. A particle can travel through space as well through time (thermodynamic time). The combined velocities form the world-line, which is developing itself as the universe unfolds.

In this talk, we explore the relation between smoothly varying couplings and Lorentz violation. Within the context of a supergravity model, we present an explicit mechanism that causes the effective fine-structure constant and the effective electromagnetic θ angle to acquire related spacetime dependences. We argue that this leads to potentially observable Lorentz violation and discuss some implications for the standard-model extension.

What if the universe had no beginning, and time stretched back infinitely without a big bang to start things off? That's one possible consequence of an idea called "rainbow gravity," so-named because it posits that gravity's effects on spacetime are felt differently by different wavelengths of light, aka different colors in the rainbow.

Rainbow gravity was first proposed 10 years ago as a possible step toward repairing the rifts between the theories of general relativity (covering the very big) and quantum mechanics (concerning the realm of the very small). The idea is not a complete theory for describing quantum effects on gravity, and is not widely accepted. Nevertheless, physicists have now applied the concept to the question of how the universe began, and found that if rainbow gravity is correct, spacetime may have a drastically different origin story than the widely accepted picture of the big bang.

According to Einstein's general relativity, massive objects warp spacetime so that anything traveling through it, including light, takes a curving path. Standard physics says this path shouldn't depend on the energy of the particles moving through spacetime, but in rainbow gravity, it does. "Particles with different energies will actually see different spacetimes, different gravitational fields," says Adel Awad of the Center for Theoretical Physics at Zewail City of Science and Technology in Egypt, who led the new research, published in October in the Journal of Cosmology and Astroparticle Physics. The color of light is determined by its frequency, and because different frequencies correspond to different energies, light particles (photons) of different colors would travel on slightly different paths though spacetime, according to their energy. 041b061a72

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