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- Puzzled by the apparent outcome of the formula.

Hi,

I'm just an amateur, no physics qualifications. I've been reading here.

Professor Richard Conn Henry likes the word trivial. None of this is trivial for me.

Einstein apparently said:

(sorry don't know how to present the maths correctly here, dx2 means 'dx squared'.)

dx2 + dy2 + dz2 - dt2 = ds2 = dx'2 + dy'2 + dz'2 -dt'2 (i.e. ds is invariant for any observer)

Professor Henry goes quickly on to show how photons know no time. I'm a bit (lot) slower than him and light.

So I'm just looking at:

(assume, as they do, the motion is only in x)

ds2 = dx2 - dt2

The German word for distance is Strecke. So German mathematicians will often use S or s for Strecke, distance, in formulas. I assume, ds is a distance, the distance from the point of origin, in whatever units, let's say metres, as are dy and dz.

Now, as light seems to have a constant speed, we can regard 1 second of time as "the length light travels in 1 second". So multiply time by, let's say c = 300 000 metres per second, as a round number, and we have a 'light distance second' in units of metres. I think that is what they are doing. Certainly, you can't just add seconds to metres.

ds2 = dx2 - cdt2

In Einstein's 4-dimensional Pythagorean type calculations I get some funny results.

Assume the velocity of the object I observe is 1 metre per second and time, dt, is 1 second, so I observe something travel 1m, dx = 1m

ds2 = 1 - 300 000 * 1 = - 299 999 metres, ds = √-299 999

Actually, I observe the object moved 1m from me in 1s and therefore ds = 1 = dx

Obviously, I don't know what I'm doing. Could you please point me right? What exactly is Einstein's ds?

I'm just an amateur, no physics qualifications. I've been reading here.

Professor Richard Conn Henry likes the word trivial. None of this is trivial for me.

Einstein apparently said:

(sorry don't know how to present the maths correctly here, dx2 means 'dx squared'.)

dx2 + dy2 + dz2 - dt2 = ds2 = dx'2 + dy'2 + dz'2 -dt'2 (i.e. ds is invariant for any observer)

Professor Henry goes quickly on to show how photons know no time. I'm a bit (lot) slower than him and light.

So I'm just looking at:

(assume, as they do, the motion is only in x)

ds2 = dx2 - dt2

The German word for distance is Strecke. So German mathematicians will often use S or s for Strecke, distance, in formulas. I assume, ds is a distance, the distance from the point of origin, in whatever units, let's say metres, as are dy and dz.

Now, as light seems to have a constant speed, we can regard 1 second of time as "the length light travels in 1 second". So multiply time by, let's say c = 300 000 metres per second, as a round number, and we have a 'light distance second' in units of metres. I think that is what they are doing. Certainly, you can't just add seconds to metres.

ds2 = dx2 - cdt2

In Einstein's 4-dimensional Pythagorean type calculations I get some funny results.

Assume the velocity of the object I observe is 1 metre per second and time, dt, is 1 second, so I observe something travel 1m, dx = 1m

ds2 = 1 - 300 000 * 1 = - 299 999 metres, ds = √-299 999

Actually, I observe the object moved 1m from me in 1s and therefore ds = 1 = dx

Obviously, I don't know what I'm doing. Could you please point me right? What exactly is Einstein's ds?