Recent content by PFStudent

Homework Statement Conventionally, the Lorentz Transformation relates two reference frames that begin at the same location and time with one reference frame moving at a constant velocity {\vec{v}} along a positive {x}-axis (which is common to both reference frames) with respect to the other...

Hey, Well, since we're beginning at, {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}} {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}} {{z}^{\prime}} = {{z}-{{{v}_{z}}{t}}} {{t}^{\prime}} = {t} We can let, {\vec{v}} = {{\vec{v}}_{\rho}} {{\vec{v}}_{r}} = {{\vec{v}}_{x}}+{{\vec{v}}_{y}}...

Hey, Any help on this, still not sure if I'm on the right path here. Thanks, -PFStudent

Hey, Well, I assume this part is right, {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}} {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}} {{z}^{\prime}} = {{x}-{{{v}_{z}}{t}}} {{t}^{\prime}} = {t} so where do I go from here? Thanks, -PFStudent

Hey, Well, in the 1-D case we are only considering motion along one axis, where {\vec{v}} is a constant, hence the following Galilean Transformation, {{x}^{\prime}} = {{x}-{vt}} {{y}^{\prime}} = {y} {{z}^{\prime}} = {z} {{t}^{\prime}} = {t} So, my guess is that by beginning with the...

Homework Statement Show that the form of Newton's Second Law is invariant under: (a). a Galilean Transformation (GT) in 1-Dimension. (b). a Galilean Transformation (GT) in 2-Dimensions. (c). a Galilean Transformation (GT) in 3-Dimensions. Homework Equations Newton's Second Law...

Hey, Ok, we can build on this and establish that the Galilean Transformation (GT) for: 1, 2, and 3; dimensions is the following, GT for 1-D {{x}^{\prime}} = {{x}-{vt}} {{y}^{\prime}} = {y} {{z}^{\prime}} = {z} {{t}^{\prime}} = {t} GT for 2-D (Converted via using Polar Coordinates...

Homework Statement Conventionally, the Galilean Transformation relates two reference frames that begin at the same location and time with one reference frame moving at a constant velocity {\vec{v}} along a positive {x}-axis (which is common to both reference frames) with respect to the other...

Hey, An interesting problem, anyone have a solution to it? Thanks, -PFStudent

Hey, A more substantial and detailed reply would help, anyone? Thanks, -PFStudent

Hey, Still stuck on these questions, a little bit of help would be nice. Thanks, -PFStudent

Hey, 1. Homework Statement . I was reading through the Differential Equation portion of my textbook and didn't quite understand the following paragraph. The above paragraph seemed a little confusing since, conventionally, {y} = {f(x)} Where, {{\frac...

Hey, Not quite, from my understanding, the displacement {{\vec{r}}_{s}} is NOT the same as the displacement {\vec{r}} used when calculating the Work done by the spring. The reason for this is because in, {{\vec{F}}_{s}} = {{-k}{{\vec{r}}_{s}}} the spring's displacement {{\vec{r}}_{s}} is...

Hey, I see. Ok, so let me see if I understand this correctly. When considering the formal equation for the Force due to a spring, {{\vec{F}}_{s}} = {{-k}{\vec{r}}} it is understood that, {\vec{r}} is defined as the displacement and, {\vec{r}} = {{\Delta}{\vec{r}}} where...

Hey, In that case how would one ever derive, {{W}_{s}} = {\frac{k}{2}\Biggl({{{{r}_{i}}^{2}}{-}{{{r}_{f}}^{2}}}\Biggr)} if we could only evaluate our integral, to find the Work done by a spring, from {0} to {{\vec{r}}_{f}}? Instead of the above, if we are limited to evaluating from...