At one poinn a pipeline the water's spced is 3.00 ms and the gange pressure is

Sx l0 Pa. Find the gauge peessure al a seoond point in the line, 11.0 m bower thas

the first, if the pipe dianeter at the seccond point is twice that at the first.

Munaza Awan
2022-07-31

At one poinn a pipeline the water's spced is 3.00 ms and the gange pressure is

Sx l0 Pa. Find the gauge peessure al a seoond point in the line, 11.0 m bower thas

the first, if the pipe dianeter at the seccond point is twice that at the first.

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asked 2022-07-14

If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?

For example:

In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V{)}^{2}$. However, another observer on the frame calculates the energy as $\frac{1}{2}m{v}^{2}$.

For example:

In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V{)}^{2}$. However, another observer on the frame calculates the energy as $\frac{1}{2}m{v}^{2}$.

asked 2022-07-16

Special relativity says that anything moving (almost) at the speed of light will look like its internal clock has (almost) stopped from the perspective of a stationary observer. How do we see light as alternating electric and magnetic fields?

asked 2022-08-24

Is there any way in which a bound state could consist only of massless particles? If yes, would this "atom" of massless particles travel on a light-like trajectory, or would the interaction energy cause it to travel on a time-like trajectory?

asked 2022-07-16

Can one explain the relativistic energy transformation formula:

$E=\gamma \text{}{E}^{\prime},$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu $?

Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity 𝑣.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma \text{}T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma \text{}T/T=\gamma $ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma $ over the frequency of A's system.

As we have the quantum relation, $E=h\nu $, this implies that B observes that the energy of his quantum system is a factor of $\gamma $ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

$E=\gamma \text{}{E}^{\prime},$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu $?

Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity 𝑣.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma \text{}T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma \text{}T/T=\gamma $ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma $ over the frequency of A's system.

As we have the quantum relation, $E=h\nu $, this implies that B observes that the energy of his quantum system is a factor of $\gamma $ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

asked 2022-04-07

Why the log? Is it there to make the growth of the function slower?

As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\mathrm{\infty})$ to $(-\mathrm{\infty},\mathrm{\infty})$ (For a particle emitted along the beam axis after collision $\theta =0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)

As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\mathrm{\infty})$ to $(-\mathrm{\infty},\mathrm{\infty})$ (For a particle emitted along the beam axis after collision $\theta =0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)

asked 2022-05-20

It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume $-+++$ signature. Given ${\gamma}_{\mu}$ as gamma matrices that satisfy $\mathrm{t}\mathrm{r}({\gamma}_{\mu}{\gamma}_{\nu})=4{\eta}_{\mu \nu}$, then we have $\mathrm{t}\mathrm{r}({\gamma}_{\mu}^{\prime}{\gamma}_{\nu}^{\prime})=4{\eta}_{\mu \nu}$ if we put:${\gamma}_{\mu}^{\prime}=\mathrm{exp}(-A){\gamma}_{\mu}\mathrm{exp}(+A)$where $A$ is any matrix.

Boosts and rotations use $A$ as a bivector. For example, with $\alpha $ a real number, $A=\alpha {\gamma}_{0}{\gamma}_{3}$ boosts in the $z$ direction while $A=\alpha {\gamma}_{1}{\gamma}_{2}$ gives a rotation around the $z$ axis.

Solving for the value of α that gives a boost with velocity $\beta =v/c$ appears to be straightforward. But how to do the rotations?

And by the way, what happens when you generalize $A$ to be something other than bivectors?

Boosts and rotations use $A$ as a bivector. For example, with $\alpha $ a real number, $A=\alpha {\gamma}_{0}{\gamma}_{3}$ boosts in the $z$ direction while $A=\alpha {\gamma}_{1}{\gamma}_{2}$ gives a rotation around the $z$ axis.

Solving for the value of α that gives a boost with velocity $\beta =v/c$ appears to be straightforward. But how to do the rotations?

And by the way, what happens when you generalize $A$ to be something other than bivectors?

asked 2022-04-07

While the speed of light in vacuum is a universal constant (c), the speed at which light propagates in other materials/mediums may be less than c. This is obviously suggested by the fact that different materials (especially in the case of transparent ones) have a particular refractive index.

But surely, matter or even photons can be accelerated beyond this speed in a medium?

But surely, matter or even photons can be accelerated beyond this speed in a medium?